xref: /dports/java/openjdk13/jdk13u-jdk-13.0.10-1-1/test/jdk/jdk/internal/math/FloatingDecimal/OldFloatingDecimalForTest.java

/*
 * Copyright (c) 1996, 2013, Oracle and/or its affiliates. All rights reserved.
 * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER.
 *
 * This code is free software; you can redistribute it and/or modify it
 * under the terms of the GNU General Public License version 2 only, as
 * published by the Free Software Foundation.
 *
 * This code is distributed in the hope that it will be useful, but WITHOUT
 * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or
 * FITNESS FOR A PARTICULAR PURPOSE.  See the GNU General Public License
 * version 2 for more details (a copy is included in the LICENSE file that
 * accompanied this code).
 *
 * You should have received a copy of the GNU General Public License version
 * 2 along with this work; if not, write to the Free Software Foundation,
 * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA.
 *
 * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA
 * or visit www.oracle.com if you need additional information or have any
 * questions.
 */

//package jdk.internal.math;

import java.util.regex.*;

public class OldFloatingDecimalForTest{
    boolean     isExceptional;
    boolean     isNegative;
    int         decExponent;
    char        digits[];
    int         nDigits;
    int         bigIntExp;
    int         bigIntNBits;
    boolean     mustSetRoundDir = false;
    boolean     fromHex = false;
    int         roundDir = 0; // set by doubleValue

    /*
     * The fields below provides additional information about the result of
     * the binary to decimal digits conversion done in dtoa() and roundup()
     * methods. They are changed if needed by those two methods.
     */

    // True if the dtoa() binary to decimal conversion was exact.
    boolean     exactDecimalConversion = false;

    // True if the result of the binary to decimal conversion was rounded-up
    // at the end of the conversion process, i.e. roundUp() method was called.
    boolean     decimalDigitsRoundedUp = false;

    private     OldFloatingDecimalForTest( boolean negSign, int decExponent, char []digits, int n,  boolean e )
    {
        isNegative = negSign;
        isExceptional = e;
        this.decExponent = decExponent;
        this.digits = digits;
        this.nDigits = n;
    }

    /*
     * Constants of the implementation
     * Most are IEEE-754 related.
     * (There are more really boring constants at the end.)
     */
    static final long   signMask = 0x8000000000000000L;
    static final long   expMask  = 0x7ff0000000000000L;
    static final long   fractMask= ~(signMask|expMask);
    static final int    expShift = 52;
    static final int    expBias  = 1023;
    static final long   fractHOB = ( 1L<<expShift ); // assumed High-Order bit
    static final long   expOne   = ((long)expBias)<<expShift; // exponent of 1.0
    static final int    maxSmallBinExp = 62;
    static final int    minSmallBinExp = -( 63 / 3 );
    static final int    maxDecimalDigits = 15;
    static final int    maxDecimalExponent = 308;
    static final int    minDecimalExponent = -324;
    static final int    bigDecimalExponent = 324; // i.e. abs(minDecimalExponent)

    static final long   highbyte = 0xff00000000000000L;
    static final long   highbit  = 0x8000000000000000L;
    static final long   lowbytes = ~highbyte;

    static final int    singleSignMask =    0x80000000;
    static final int    singleExpMask  =    0x7f800000;
    static final int    singleFractMask =   ~(singleSignMask|singleExpMask);
    static final int    singleExpShift  =   23;
    static final int    singleFractHOB  =   1<<singleExpShift;
    static final int    singleExpBias   =   127;
    static final int    singleMaxDecimalDigits = 7;
    static final int    singleMaxDecimalExponent = 38;
    static final int    singleMinDecimalExponent = -45;

    static final int    intDecimalDigits = 9;


    /*
     * count number of bits from high-order 1 bit to low-order 1 bit,
     * inclusive.
     */
    private static int
    countBits( long v ){
        //
        // the strategy is to shift until we get a non-zero sign bit
        // then shift until we have no bits left, counting the difference.
        // we do byte shifting as a hack. Hope it helps.
        //
        if ( v == 0L ) return 0;

        while ( ( v & highbyte ) == 0L ){
            v <<= 8;
        }
        while ( v > 0L ) { // i.e. while ((v&highbit) == 0L )
            v <<= 1;
        }

        int n = 0;
        while (( v & lowbytes ) != 0L ){
            v <<= 8;
            n += 8;
        }
        while ( v != 0L ){
            v <<= 1;
            n += 1;
        }
        return n;
    }

    /*
     * Keep big powers of 5 handy for future reference.
     */
    private static OldFDBigIntForTest b5p[];

    private static synchronized OldFDBigIntForTest
    big5pow( int p ){
        assert p >= 0 : p; // negative power of 5
        if ( b5p == null ){
            b5p = new OldFDBigIntForTest[ p+1 ];
        }else if (b5p.length <= p ){
            OldFDBigIntForTest t[] = new OldFDBigIntForTest[ p+1 ];
            System.arraycopy( b5p, 0, t, 0, b5p.length );
            b5p = t;
        }
        if ( b5p[p] != null )
            return b5p[p];
        else if ( p < small5pow.length )
            return b5p[p] = new OldFDBigIntForTest( small5pow[p] );
        else if ( p < long5pow.length )
            return b5p[p] = new OldFDBigIntForTest( long5pow[p] );
        else {
            // construct the value.
            // recursively.
            int q, r;
            // in order to compute 5^p,
            // compute its square root, 5^(p/2) and square.
            // or, let q = p / 2, r = p -q, then
            // 5^p = 5^(q+r) = 5^q * 5^r
            q = p >> 1;
            r = p - q;
            OldFDBigIntForTest bigq =  b5p[q];
            if ( bigq == null )
                bigq = big5pow ( q );
            if ( r < small5pow.length ){
                return (b5p[p] = bigq.mult( small5pow[r] ) );
            }else{
                OldFDBigIntForTest bigr = b5p[ r ];
                if ( bigr == null )
                    bigr = big5pow( r );
                return (b5p[p] = bigq.mult( bigr ) );
            }
        }
    }

    //
    // a common operation
    //
    private static OldFDBigIntForTest
    multPow52( OldFDBigIntForTest v, int p5, int p2 ){
        if ( p5 != 0 ){
            if ( p5 < small5pow.length ){
                v = v.mult( small5pow[p5] );
            } else {
                v = v.mult( big5pow( p5 ) );
            }
        }
        if ( p2 != 0 ){
            v.lshiftMe( p2 );
        }
        return v;
    }

    //
    // another common operation
    //
    private static OldFDBigIntForTest
    constructPow52( int p5, int p2 ){
        OldFDBigIntForTest v = new OldFDBigIntForTest( big5pow( p5 ) );
        if ( p2 != 0 ){
            v.lshiftMe( p2 );
        }
        return v;
    }

    /*
     * Make a floating double into a OldFDBigIntForTest.
     * This could also be structured as a OldFDBigIntForTest
     * constructor, but we'd have to build a lot of knowledge
     * about floating-point representation into it, and we don't want to.
     *
     * AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES
     * bigIntExp and bigIntNBits
     *
     */
    private OldFDBigIntForTest
    doubleToBigInt( double dval ){
        long lbits = Double.doubleToLongBits( dval ) & ~signMask;
        int binexp = (int)(lbits >>> expShift);
        lbits &= fractMask;
        if ( binexp > 0 ){
            lbits |= fractHOB;
        } else {
            assert lbits != 0L : lbits; // doubleToBigInt(0.0)
            binexp +=1;
            while ( (lbits & fractHOB ) == 0L){
                lbits <<= 1;
                binexp -= 1;
            }
        }
        binexp -= expBias;
        int nbits = countBits( lbits );
        /*
         * We now know where the high-order 1 bit is,
         * and we know how many there are.
         */
        int lowOrderZeros = expShift+1-nbits;
        lbits >>>= lowOrderZeros;

        bigIntExp = binexp+1-nbits;
        bigIntNBits = nbits;
        return new OldFDBigIntForTest( lbits );
    }

    /*
     * Compute a number that is the ULP of the given value,
     * for purposes of addition/subtraction. Generally easy.
     * More difficult if subtracting and the argument
     * is a normalized a power of 2, as the ULP changes at these points.
     */
    private static double ulp( double dval, boolean subtracting ){
        long lbits = Double.doubleToLongBits( dval ) & ~signMask;
        int binexp = (int)(lbits >>> expShift);
        double ulpval;
        if ( subtracting && ( binexp >= expShift ) && ((lbits&fractMask) == 0L) ){
            // for subtraction from normalized, powers of 2,
            // use next-smaller exponent
            binexp -= 1;
        }
        if ( binexp > expShift ){
            ulpval = Double.longBitsToDouble( ((long)(binexp-expShift))<<expShift );
        } else if ( binexp == 0 ){
            ulpval = Double.MIN_VALUE;
        } else {
            ulpval = Double.longBitsToDouble( 1L<<(binexp-1) );
        }
        if ( subtracting ) ulpval = - ulpval;

        return ulpval;
    }

    /*
     * Round a double to a float.
     * In addition to the fraction bits of the double,
     * look at the class instance variable roundDir,
     * which should help us avoid double-rounding error.
     * roundDir was set in hardValueOf if the estimate was
     * close enough, but not exact. It tells us which direction
     * of rounding is preferred.
     */
    float
    stickyRound( double dval ){
        long lbits = Double.doubleToLongBits( dval );
        long binexp = lbits & expMask;
        if ( binexp == 0L || binexp == expMask ){
            // what we have here is special.
            // don't worry, the right thing will happen.
            return (float) dval;
        }
        lbits += (long)roundDir; // hack-o-matic.
        return (float)Double.longBitsToDouble( lbits );
    }


    /*
     * This is the easy subcase --
     * all the significant bits, after scaling, are held in lvalue.
     * negSign and decExponent tell us what processing and scaling
     * has already been done. Exceptional cases have already been
     * stripped out.
     * In particular:
     * lvalue is a finite number (not Inf, nor NaN)
     * lvalue > 0L (not zero, nor negative).
     *
     * The only reason that we develop the digits here, rather than
     * calling on Long.toString() is that we can do it a little faster,
     * and besides want to treat trailing 0s specially. If Long.toString
     * changes, we should re-evaluate this strategy!
     */
    private void
    developLongDigits( int decExponent, long lvalue, long insignificant ){
        char digits[];
        int  ndigits;
        int  digitno;
        int  c;
        //
        // Discard non-significant low-order bits, while rounding,
        // up to insignificant value.
        int i;
        for ( i = 0; insignificant >= 10L; i++ )
            insignificant /= 10L;
        if ( i != 0 ){
            long pow10 = long5pow[i] << i; // 10^i == 5^i * 2^i;
            long residue = lvalue % pow10;
            lvalue /= pow10;
            decExponent += i;
            if ( residue >= (pow10>>1) ){
                // round up based on the low-order bits we're discarding
                lvalue++;
            }
        }
        if ( lvalue <= Integer.MAX_VALUE ){
            assert lvalue > 0L : lvalue; // lvalue <= 0
            // even easier subcase!
            // can do int arithmetic rather than long!
            int  ivalue = (int)lvalue;
            ndigits = 10;
            digits = perThreadBuffer.get();
            digitno = ndigits-1;
            c = ivalue%10;
            ivalue /= 10;
            while ( c == 0 ){
                decExponent++;
                c = ivalue%10;
                ivalue /= 10;
            }
            while ( ivalue != 0){
                digits[digitno--] = (char)(c+'0');
                decExponent++;
                c = ivalue%10;
                ivalue /= 10;
            }
            digits[digitno] = (char)(c+'0');
        } else {
            // same algorithm as above (same bugs, too )
            // but using long arithmetic.
            ndigits = 20;
            digits = perThreadBuffer.get();
            digitno = ndigits-1;
            c = (int)(lvalue%10L);
            lvalue /= 10L;
            while ( c == 0 ){
                decExponent++;
                c = (int)(lvalue%10L);
                lvalue /= 10L;
            }
            while ( lvalue != 0L ){
                digits[digitno--] = (char)(c+'0');
                decExponent++;
                c = (int)(lvalue%10L);
                lvalue /= 10;
            }
            digits[digitno] = (char)(c+'0');
        }
        char result [];
        ndigits -= digitno;
        result = new char[ ndigits ];
        System.arraycopy( digits, digitno, result, 0, ndigits );
        this.digits = result;
        this.decExponent = decExponent+1;
        this.nDigits = ndigits;
    }

    //
    // add one to the least significant digit.
    // in the unlikely event there is a carry out,
    // deal with it.
    // assert that this will only happen where there
    // is only one digit, e.g. (float)1e-44 seems to do it.
    //
    private void
    roundup(){
        int i;
        int q = digits[ i = (nDigits-1)];
        if ( q == '9' ){
            while ( q == '9' && i > 0 ){
                digits[i] = '0';
                q = digits[--i];
            }
            if ( q == '9' ){
                // carryout! High-order 1, rest 0s, larger exp.
                decExponent += 1;
                digits[0] = '1';
                return;
            }
            // else fall through.
        }
        digits[i] = (char)(q+1);
        decimalDigitsRoundedUp = true;
    }

    public boolean digitsRoundedUp() {
        return decimalDigitsRoundedUp;
    }

    /*
     * FIRST IMPORTANT CONSTRUCTOR: DOUBLE
     */
    public OldFloatingDecimalForTest( double d )
    {
        long    dBits = Double.doubleToLongBits( d );
        long    fractBits;
        int     binExp;
        int     nSignificantBits;

        // discover and delete sign
        if ( (dBits&signMask) != 0 ){
            isNegative = true;
            dBits ^= signMask;
        } else {
            isNegative = false;
        }
        // Begin to unpack
        // Discover obvious special cases of NaN and Infinity.
        binExp = (int)( (dBits&expMask) >> expShift );
        fractBits = dBits&fractMask;
        if ( binExp == (int)(expMask>>expShift) ) {
            isExceptional = true;
            if ( fractBits == 0L ){
                digits =  infinity;
            } else {
                digits = notANumber;
                isNegative = false; // NaN has no sign!
            }
            nDigits = digits.length;
            return;
        }
        isExceptional = false;
        // Finish unpacking
        // Normalize denormalized numbers.
        // Insert assumed high-order bit for normalized numbers.
        // Subtract exponent bias.
        if ( binExp == 0 ){
            if ( fractBits == 0L ){
                // not a denorm, just a 0!
                decExponent = 0;
                digits = zero;
                nDigits = 1;
                return;
            }
            while ( (fractBits&fractHOB) == 0L ){
                fractBits <<= 1;
                binExp -= 1;
            }
            nSignificantBits = expShift + binExp +1; // recall binExp is  - shift count.
            binExp += 1;
        } else {
            fractBits |= fractHOB;
            nSignificantBits = expShift+1;
        }
        binExp -= expBias;
        // call the routine that actually does all the hard work.
        dtoa( binExp, fractBits, nSignificantBits );
    }

    /*
     * SECOND IMPORTANT CONSTRUCTOR: SINGLE
     */
    public OldFloatingDecimalForTest( float f )
    {
        int     fBits = Float.floatToIntBits( f );
        int     fractBits;
        int     binExp;
        int     nSignificantBits;

        // discover and delete sign
        if ( (fBits&singleSignMask) != 0 ){
            isNegative = true;
            fBits ^= singleSignMask;
        } else {
            isNegative = false;
        }
        // Begin to unpack
        // Discover obvious special cases of NaN and Infinity.
        binExp = (fBits&singleExpMask) >> singleExpShift;
        fractBits = fBits&singleFractMask;
        if ( binExp == (singleExpMask>>singleExpShift) ) {
            isExceptional = true;
            if ( fractBits == 0L ){
                digits =  infinity;
            } else {
                digits = notANumber;
                isNegative = false; // NaN has no sign!
            }
            nDigits = digits.length;
            return;
        }
        isExceptional = false;
        // Finish unpacking
        // Normalize denormalized numbers.
        // Insert assumed high-order bit for normalized numbers.
        // Subtract exponent bias.
        if ( binExp == 0 ){
            if ( fractBits == 0 ){
                // not a denorm, just a 0!
                decExponent = 0;
                digits = zero;
                nDigits = 1;
                return;
            }
            while ( (fractBits&singleFractHOB) == 0 ){
                fractBits <<= 1;
                binExp -= 1;
            }
            nSignificantBits = singleExpShift + binExp +1; // recall binExp is  - shift count.
            binExp += 1;
        } else {
            fractBits |= singleFractHOB;
            nSignificantBits = singleExpShift+1;
        }
        binExp -= singleExpBias;
        // call the routine that actually does all the hard work.
        dtoa( binExp, ((long)fractBits)<<(expShift-singleExpShift), nSignificantBits );
    }

    private void
    dtoa( int binExp, long fractBits, int nSignificantBits )
    {
        int     nFractBits; // number of significant bits of fractBits;
        int     nTinyBits;  // number of these to the right of the point.
        int     decExp;

        // Examine number. Determine if it is an easy case,
        // which we can do pretty trivially using float/long conversion,
        // or whether we must do real work.
        nFractBits = countBits( fractBits );
        nTinyBits = Math.max( 0, nFractBits - binExp - 1 );
        if ( binExp <= maxSmallBinExp && binExp >= minSmallBinExp ){
            // Look more closely at the number to decide if,
            // with scaling by 10^nTinyBits, the result will fit in
            // a long.
            if ( (nTinyBits < long5pow.length) && ((nFractBits + n5bits[nTinyBits]) < 64 ) ){
                /*
                 * We can do this:
                 * take the fraction bits, which are normalized.
                 * (a) nTinyBits == 0: Shift left or right appropriately
                 *     to align the binary point at the extreme right, i.e.
                 *     where a long int point is expected to be. The integer
                 *     result is easily converted to a string.
                 * (b) nTinyBits > 0: Shift right by expShift-nFractBits,
                 *     which effectively converts to long and scales by
                 *     2^nTinyBits. Then multiply by 5^nTinyBits to
                 *     complete the scaling. We know this won't overflow
                 *     because we just counted the number of bits necessary
                 *     in the result. The integer you get from this can
                 *     then be converted to a string pretty easily.
                 */
                long halfULP;
                if ( nTinyBits == 0 ) {
                    if ( binExp > nSignificantBits ){
                        halfULP = 1L << ( binExp-nSignificantBits-1);
                    } else {
                        halfULP = 0L;
                    }
                    if ( binExp >= expShift ){
                        fractBits <<= (binExp-expShift);
                    } else {
                        fractBits >>>= (expShift-binExp) ;
                    }
                    developLongDigits( 0, fractBits, halfULP );
                    return;
                }
                /*
                 * The following causes excess digits to be printed
                 * out in the single-float case. Our manipulation of
                 * halfULP here is apparently not correct. If we
                 * better understand how this works, perhaps we can
                 * use this special case again. But for the time being,
                 * we do not.
                 * else {
                 *     fractBits >>>= expShift+1-nFractBits;
                 *     fractBits *= long5pow[ nTinyBits ];
                 *     halfULP = long5pow[ nTinyBits ] >> (1+nSignificantBits-nFractBits);
                 *     developLongDigits( -nTinyBits, fractBits, halfULP );
                 *     return;
                 * }
                 */
            }
        }
        /*
         * This is the hard case. We are going to compute large positive
         * integers B and S and integer decExp, s.t.
         *      d = ( B / S ) * 10^decExp
         *      1 <= B / S < 10
         * Obvious choices are:
         *      decExp = floor( log10(d) )
         *      B      = d * 2^nTinyBits * 10^max( 0, -decExp )
         *      S      = 10^max( 0, decExp) * 2^nTinyBits
         * (noting that nTinyBits has already been forced to non-negative)
         * I am also going to compute a large positive integer
         *      M      = (1/2^nSignificantBits) * 2^nTinyBits * 10^max( 0, -decExp )
         * i.e. M is (1/2) of the ULP of d, scaled like B.
         * When we iterate through dividing B/S and picking off the
         * quotient bits, we will know when to stop when the remainder
         * is <= M.
         *
         * We keep track of powers of 2 and powers of 5.
         */

        /*
         * Estimate decimal exponent. (If it is small-ish,
         * we could double-check.)
         *
         * First, scale the mantissa bits such that 1 <= d2 < 2.
         * We are then going to estimate
         *          log10(d2) ~=~  (d2-1.5)/1.5 + log(1.5)
         * and so we can estimate
         *      log10(d) ~=~ log10(d2) + binExp * log10(2)
         * take the floor and call it decExp.
         * FIXME -- use more precise constants here. It costs no more.
         */
        double d2 = Double.longBitsToDouble(
            expOne | ( fractBits &~ fractHOB ) );
        decExp = (int)Math.floor(
            (d2-1.5D)*0.289529654D + 0.176091259 + (double)binExp * 0.301029995663981 );
        int B2, B5; // powers of 2 and powers of 5, respectively, in B
        int S2, S5; // powers of 2 and powers of 5, respectively, in S
        int M2, M5; // powers of 2 and powers of 5, respectively, in M
        int Bbits; // binary digits needed to represent B, approx.
        int tenSbits; // binary digits needed to represent 10*S, approx.
        OldFDBigIntForTest Sval, Bval, Mval;

        B5 = Math.max( 0, -decExp );
        B2 = B5 + nTinyBits + binExp;

        S5 = Math.max( 0, decExp );
        S2 = S5 + nTinyBits;

        M5 = B5;
        M2 = B2 - nSignificantBits;

        /*
         * the long integer fractBits contains the (nFractBits) interesting
         * bits from the mantissa of d ( hidden 1 added if necessary) followed
         * by (expShift+1-nFractBits) zeros. In the interest of compactness,
         * I will shift out those zeros before turning fractBits into a
         * OldFDBigIntForTest. The resulting whole number will be
         *      d * 2^(nFractBits-1-binExp).
         */
        fractBits >>>= (expShift+1-nFractBits);
        B2 -= nFractBits-1;
        int common2factor = Math.min( B2, S2 );
        B2 -= common2factor;
        S2 -= common2factor;
        M2 -= common2factor;

        /*
         * HACK!! For exact powers of two, the next smallest number
         * is only half as far away as we think (because the meaning of
         * ULP changes at power-of-two bounds) for this reason, we
         * hack M2. Hope this works.
         */
        if ( nFractBits == 1 )
            M2 -= 1;

        if ( M2 < 0 ){
            // oops.
            // since we cannot scale M down far enough,
            // we must scale the other values up.
            B2 -= M2;
            S2 -= M2;
            M2 =  0;
        }
        /*
         * Construct, Scale, iterate.
         * Some day, we'll write a stopping test that takes
         * account of the asymmetry of the spacing of floating-point
         * numbers below perfect powers of 2
         * 26 Sept 96 is not that day.
         * So we use a symmetric test.
         */
        char digits[] = this.digits = new char[18];
        int  ndigit = 0;
        boolean low, high;
        long lowDigitDifference;
        int  q;

        /*
         * Detect the special cases where all the numbers we are about
         * to compute will fit in int or long integers.
         * In these cases, we will avoid doing OldFDBigIntForTest arithmetic.
         * We use the same algorithms, except that we "normalize"
         * our OldFDBigIntForTests before iterating. This is to make division easier,
         * as it makes our fist guess (quotient of high-order words)
         * more accurate!
         *
         * Some day, we'll write a stopping test that takes
         * account of the asymmetry of the spacing of floating-point
         * numbers below perfect powers of 2
         * 26 Sept 96 is not that day.
         * So we use a symmetric test.
         */
        Bbits = nFractBits + B2 + (( B5 < n5bits.length )? n5bits[B5] : ( B5*3 ));
        tenSbits = S2+1 + (( (S5+1) < n5bits.length )? n5bits[(S5+1)] : ( (S5+1)*3 ));
        if ( Bbits < 64 && tenSbits < 64){
            if ( Bbits < 32 && tenSbits < 32){
                // wa-hoo! They're all ints!
                int b = ((int)fractBits * small5pow[B5] ) << B2;
                int s = small5pow[S5] << S2;
                int m = small5pow[M5] << M2;
                int tens = s * 10;
                /*
                 * Unroll the first iteration. If our decExp estimate
                 * was too high, our first quotient will be zero. In this
                 * case, we discard it and decrement decExp.
                 */
                ndigit = 0;
                q = b / s;
                b = 10 * ( b % s );
                m *= 10;
                low  = (b <  m );
                high = (b+m > tens );
                assert q < 10 : q; // excessively large digit
                if ( (q == 0) && ! high ){
                    // oops. Usually ignore leading zero.
                    decExp--;
                } else {
                    digits[ndigit++] = (char)('0' + q);
                }
                /*
                 * HACK! Java spec sez that we always have at least
                 * one digit after the . in either F- or E-form output.
                 * Thus we will need more than one digit if we're using
                 * E-form
                 */
                if ( decExp < -3 || decExp >= 8 ){
                    high = low = false;
                }
                while( ! low && ! high ){
                    q = b / s;
                    b = 10 * ( b % s );
                    m *= 10;
                    assert q < 10 : q; // excessively large digit
                    if ( m > 0L ){
                        low  = (b <  m );
                        high = (b+m > tens );
                    } else {
                        // hack -- m might overflow!
                        // in this case, it is certainly > b,
                        // which won't
                        // and b+m > tens, too, since that has overflowed
                        // either!
                        low = true;
                        high = true;
                    }
                    digits[ndigit++] = (char)('0' + q);
                }
                lowDigitDifference = (b<<1) - tens;
                exactDecimalConversion  = (b == 0);
            } else {
                // still good! they're all longs!
                long b = (fractBits * long5pow[B5] ) << B2;
                long s = long5pow[S5] << S2;
                long m = long5pow[M5] << M2;
                long tens = s * 10L;
                /*
                 * Unroll the first iteration. If our decExp estimate
                 * was too high, our first quotient will be zero. In this
                 * case, we discard it and decrement decExp.
                 */
                ndigit = 0;
                q = (int) ( b / s );
                b = 10L * ( b % s );
                m *= 10L;
                low  = (b <  m );
                high = (b+m > tens );
                assert q < 10 : q; // excessively large digit
                if ( (q == 0) && ! high ){
                    // oops. Usually ignore leading zero.
                    decExp--;
                } else {
                    digits[ndigit++] = (char)('0' + q);
                }
                /*
                 * HACK! Java spec sez that we always have at least
                 * one digit after the . in either F- or E-form output.
                 * Thus we will need more than one digit if we're using
                 * E-form
                 */
                if ( decExp < -3 || decExp >= 8 ){
                    high = low = false;
                }
                while( ! low && ! high ){
                    q = (int) ( b / s );
                    b = 10 * ( b % s );
                    m *= 10;
                    assert q < 10 : q;  // excessively large digit
                    if ( m > 0L ){
                        low  = (b <  m );
                        high = (b+m > tens );
                    } else {
                        // hack -- m might overflow!
                        // in this case, it is certainly > b,
                        // which won't
                        // and b+m > tens, too, since that has overflowed
                        // either!
                        low = true;
                        high = true;
                    }
                    digits[ndigit++] = (char)('0' + q);
                }
                lowDigitDifference = (b<<1) - tens;
                exactDecimalConversion  = (b == 0);
            }
        } else {
            OldFDBigIntForTest ZeroVal = new OldFDBigIntForTest(0);
            OldFDBigIntForTest tenSval;
            int  shiftBias;

            /*
             * We really must do OldFDBigIntForTest arithmetic.
             * Fist, construct our OldFDBigIntForTest initial values.
             */
            Bval = multPow52( new OldFDBigIntForTest( fractBits  ), B5, B2 );
            Sval = constructPow52( S5, S2 );
            Mval = constructPow52( M5, M2 );


            // normalize so that division works better
            Bval.lshiftMe( shiftBias = Sval.normalizeMe() );
            Mval.lshiftMe( shiftBias );
            tenSval = Sval.mult( 10 );
            /*
             * Unroll the first iteration. If our decExp estimate
             * was too high, our first quotient will be zero. In this
             * case, we discard it and decrement decExp.
             */
            ndigit = 0;
            q = Bval.quoRemIteration( Sval );
            Mval = Mval.mult( 10 );
            low  = (Bval.cmp( Mval ) < 0);
            high = (Bval.add( Mval ).cmp( tenSval ) > 0 );
            assert q < 10 : q; // excessively large digit
            if ( (q == 0) && ! high ){
                // oops. Usually ignore leading zero.
                decExp--;
            } else {
                digits[ndigit++] = (char)('0' + q);
            }
            /*
             * HACK! Java spec sez that we always have at least
             * one digit after the . in either F- or E-form output.
             * Thus we will need more than one digit if we're using
             * E-form
             */
            if ( decExp < -3 || decExp >= 8 ){
                high = low = false;
            }
            while( ! low && ! high ){
                q = Bval.quoRemIteration( Sval );
                Mval = Mval.mult( 10 );
                assert q < 10 : q;  // excessively large digit
                low  = (Bval.cmp( Mval ) < 0);
                high = (Bval.add( Mval ).cmp( tenSval ) > 0 );
                digits[ndigit++] = (char)('0' + q);
            }
            if ( high && low ){
                Bval.lshiftMe(1);
                lowDigitDifference = Bval.cmp(tenSval);
            } else {
                lowDigitDifference = 0L; // this here only for flow analysis!
            }
            exactDecimalConversion  = (Bval.cmp( ZeroVal ) == 0);
        }
        this.decExponent = decExp+1;
        this.digits = digits;
        this.nDigits = ndigit;
        /*
         * Last digit gets rounded based on stopping condition.
         */
        if ( high ){
            if ( low ){
                if ( lowDigitDifference == 0L ){
                    // it's a tie!
                    // choose based on which digits we like.
                    if ( (digits[nDigits-1]&1) != 0 ) roundup();
                } else if ( lowDigitDifference > 0 ){
                    roundup();
                }
            } else {
                roundup();
            }
        }
    }

    public boolean decimalDigitsExact() {
        return exactDecimalConversion;
    }

    public String
    toString(){
        // most brain-dead version
        StringBuffer result = new StringBuffer( nDigits+8 );
        if ( isNegative ){ result.append( '-' ); }
        if ( isExceptional ){
            result.append( digits, 0, nDigits );
        } else {
            result.append( "0.");
            result.append( digits, 0, nDigits );
            result.append('e');
            result.append( decExponent );
        }
        return new String(result);
    }

    public String toJavaFormatString() {
        char result[] = perThreadBuffer.get();
        int i = getChars(result);
        return new String(result, 0, i);
    }

    private int getChars(char[] result) {
        assert nDigits <= 19 : nDigits; // generous bound on size of nDigits
        int i = 0;
        if (isNegative) { result[0] = '-'; i = 1; }
        if (isExceptional) {
            System.arraycopy(digits, 0, result, i, nDigits);
            i += nDigits;
        } else {
            if (decExponent > 0 && decExponent < 8) {
                // print digits.digits.
                int charLength = Math.min(nDigits, decExponent);
                System.arraycopy(digits, 0, result, i, charLength);
                i += charLength;
                if (charLength < decExponent) {
                    charLength = decExponent-charLength;
                    System.arraycopy(zero, 0, result, i, charLength);
                    i += charLength;
                    result[i++] = '.';
                    result[i++] = '0';
                } else {
                    result[i++] = '.';
                    if (charLength < nDigits) {
                        int t = nDigits - charLength;
                        System.arraycopy(digits, charLength, result, i, t);
                        i += t;
                    } else {
                        result[i++] = '0';
                    }
                }
            } else if (decExponent <=0 && decExponent > -3) {
                result[i++] = '0';
                result[i++] = '.';
                if (decExponent != 0) {
                    System.arraycopy(zero, 0, result, i, -decExponent);
                    i -= decExponent;
                }
                System.arraycopy(digits, 0, result, i, nDigits);
                i += nDigits;
            } else {
                result[i++] = digits[0];
                result[i++] = '.';
                if (nDigits > 1) {
                    System.arraycopy(digits, 1, result, i, nDigits-1);
                    i += nDigits-1;
                } else {
                    result[i++] = '0';
                }
                result[i++] = 'E';
                int e;
                if (decExponent <= 0) {
                    result[i++] = '-';
                    e = -decExponent+1;
                } else {
                    e = decExponent-1;
                }
                // decExponent has 1, 2, or 3, digits
                if (e <= 9) {
                    result[i++] = (char)(e+'0');
                } else if (e <= 99) {
                    result[i++] = (char)(e/10 +'0');
                    result[i++] = (char)(e%10 + '0');
                } else {
                    result[i++] = (char)(e/100+'0');
                    e %= 100;
                    result[i++] = (char)(e/10+'0');
                    result[i++] = (char)(e%10 + '0');
                }
            }
        }
        return i;
    }

    // Per-thread buffer for string/stringbuffer conversion
    private static ThreadLocal<char[]> perThreadBuffer = new ThreadLocal<char[]>() {
            protected synchronized char[] initialValue() {
                return new char[26];
            }
        };

    public void appendTo(Appendable buf) {
          char result[] = perThreadBuffer.get();
          int i = getChars(result);
        if (buf instanceof StringBuilder)
            ((StringBuilder) buf).append(result, 0, i);
        else if (buf instanceof StringBuffer)
            ((StringBuffer) buf).append(result, 0, i);
        else
            assert false;
    }

    @SuppressWarnings("fallthrough")
    public static OldFloatingDecimalForTest
    readJavaFormatString( String in ) throws NumberFormatException {
        boolean isNegative = false;
        boolean signSeen   = false;
        int     decExp;
        char    c;

    parseNumber:
        try{
            in = in.trim(); // don't fool around with white space.
                            // throws NullPointerException if null
            int l = in.length();
            if ( l == 0 ) throw new NumberFormatException("empty String");
            int i = 0;
            switch ( c = in.charAt( i ) ){
            case '-':
                isNegative = true;
                //FALLTHROUGH
            case '+':
                i++;
                signSeen = true;
            }

            // Check for NaN and Infinity strings
            c = in.charAt(i);
            if(c == 'N' || c == 'I') { // possible NaN or infinity
                boolean potentialNaN = false;
                char targetChars[] = null;  // char array of "NaN" or "Infinity"

                if(c == 'N') {
                    targetChars = notANumber;
                    potentialNaN = true;
                } else {
                    targetChars = infinity;
                }

                // compare Input string to "NaN" or "Infinity"
                int j = 0;
                while(i < l && j < targetChars.length) {
                    if(in.charAt(i) == targetChars[j]) {
                        i++; j++;
                    }
                    else // something is amiss, throw exception
                        break parseNumber;
                }

                // For the candidate string to be a NaN or infinity,
                // all characters in input string and target char[]
                // must be matched ==> j must equal targetChars.length
                // and i must equal l
                if( (j == targetChars.length) && (i == l) ) { // return NaN or infinity
                    return (potentialNaN ? new OldFloatingDecimalForTest(Double.NaN) // NaN has no sign
                            : new OldFloatingDecimalForTest(isNegative?
                                                  Double.NEGATIVE_INFINITY:
                                                  Double.POSITIVE_INFINITY)) ;
                }
                else { // something went wrong, throw exception
                    break parseNumber;
                }

            } else if (c == '0')  { // check for hexadecimal floating-point number
                if (l > i+1 ) {
                    char ch = in.charAt(i+1);
                    if (ch == 'x' || ch == 'X' ) // possible hex string
                        return parseHexString(in);
                }
            }  // look for and process decimal floating-point string

            char[] digits = new char[ l ];
            int    nDigits= 0;
            boolean decSeen = false;
            int decPt = 0;
            int nLeadZero = 0;
            int nTrailZero= 0;
        digitLoop:
            while ( i < l ){
                switch ( c = in.charAt( i ) ){
                case '0':
                    if ( nDigits > 0 ){
                        nTrailZero += 1;
                    } else {
                        nLeadZero += 1;
                    }
                    break; // out of switch.
                case '1':
                case '2':
                case '3':
                case '4':
                case '5':
                case '6':
                case '7':
                case '8':
                case '9':
                    while ( nTrailZero > 0 ){
                        digits[nDigits++] = '0';
                        nTrailZero -= 1;
                    }
                    digits[nDigits++] = c;
                    break; // out of switch.
                case '.':
                    if ( decSeen ){
                        // already saw one ., this is the 2nd.
                        throw new NumberFormatException("multiple points");
                    }
                    decPt = i;
                    if ( signSeen ){
                        decPt -= 1;
                    }
                    decSeen = true;
                    break; // out of switch.
                default:
                    break digitLoop;
                }
                i++;
            }
            /*
             * At this point, we've scanned all the digits and decimal
             * point we're going to see. Trim off leading and trailing
             * zeros, which will just confuse us later, and adjust
             * our initial decimal exponent accordingly.
             * To review:
             * we have seen i total characters.
             * nLeadZero of them were zeros before any other digits.
             * nTrailZero of them were zeros after any other digits.
             * if ( decSeen ), then a . was seen after decPt characters
             * ( including leading zeros which have been discarded )
             * nDigits characters were neither lead nor trailing
             * zeros, nor point
             */
            /*
             * special hack: if we saw no non-zero digits, then the
             * answer is zero!
             * Unfortunately, we feel honor-bound to keep parsing!
             */
            if ( nDigits == 0 ){
                digits = zero;
                nDigits = 1;
                if ( nLeadZero == 0 ){
                    // we saw NO DIGITS AT ALL,
                    // not even a crummy 0!
                    // this is not allowed.
                    break parseNumber; // go throw exception
                }

            }

            /* Our initial exponent is decPt, adjusted by the number of
             * discarded zeros. Or, if there was no decPt,
             * then its just nDigits adjusted by discarded trailing zeros.
             */
            if ( decSeen ){
                decExp = decPt - nLeadZero;
            } else {
                decExp = nDigits+nTrailZero;
            }

            /*
             * Look for 'e' or 'E' and an optionally signed integer.
             */
            if ( (i < l) &&  (((c = in.charAt(i) )=='e') || (c == 'E') ) ){
                int expSign = 1;
                int expVal  = 0;
                int reallyBig = Integer.MAX_VALUE / 10;
                boolean expOverflow = false;
                switch( in.charAt(++i) ){
                case '-':
                    expSign = -1;
                    //FALLTHROUGH
                case '+':
                    i++;
                }
                int expAt = i;
            expLoop:
                while ( i < l  ){
                    if ( expVal >= reallyBig ){
                        // the next character will cause integer
                        // overflow.
                        expOverflow = true;
                    }
                    switch ( c = in.charAt(i++) ){
                    case '0':
                    case '1':
                    case '2':
                    case '3':
                    case '4':
                    case '5':
                    case '6':
                    case '7':
                    case '8':
                    case '9':
                        expVal = expVal*10 + ( (int)c - (int)'0' );
                        continue;
                    default:
                        i--;           // back up.
                        break expLoop; // stop parsing exponent.
                    }
                }
                int expLimit = bigDecimalExponent+nDigits+nTrailZero;
                if ( expOverflow || ( expVal > expLimit ) ){
                    //
                    // The intent here is to end up with
                    // infinity or zero, as appropriate.
                    // The reason for yielding such a small decExponent,
                    // rather than something intuitive such as
                    // expSign*Integer.MAX_VALUE, is that this value
                    // is subject to further manipulation in
                    // doubleValue() and floatValue(), and I don't want
                    // it to be able to cause overflow there!
                    // (The only way we can get into trouble here is for
                    // really outrageous nDigits+nTrailZero, such as 2 billion. )
                    //
                    decExp = expSign*expLimit;
                } else {
                    // this should not overflow, since we tested
                    // for expVal > (MAX+N), where N >= abs(decExp)
                    decExp = decExp + expSign*expVal;
                }

                // if we saw something not a digit ( or end of string )
                // after the [Ee][+-], without seeing any digits at all
                // this is certainly an error. If we saw some digits,
                // but then some trailing garbage, that might be ok.
                // so we just fall through in that case.
                // HUMBUG
                if ( i == expAt )
                    break parseNumber; // certainly bad
            }
            /*
             * We parsed everything we could.
             * If there are leftovers, then this is not good input!
             */
            if ( i < l &&
                ((i != l - 1) ||
                (in.charAt(i) != 'f' &&
                 in.charAt(i) != 'F' &&
                 in.charAt(i) != 'd' &&
                 in.charAt(i) != 'D'))) {
                break parseNumber; // go throw exception
            }

            return new OldFloatingDecimalForTest( isNegative, decExp, digits, nDigits,  false );
        } catch ( StringIndexOutOfBoundsException e ){ }
        throw new NumberFormatException("For input string: \"" + in + "\"");
    }

    /*
     * Take a FloatingDecimal, which we presumably just scanned in,
     * and find out what its value is, as a double.
     *
     * AS A SIDE EFFECT, SET roundDir TO INDICATE PREFERRED
     * ROUNDING DIRECTION in case the result is really destined
     * for a single-precision float.
     */

    public strictfp double doubleValue(){
        int     kDigits = Math.min( nDigits, maxDecimalDigits+1 );
        long    lValue;
        double  dValue;
        double  rValue, tValue;

        // First, check for NaN and Infinity values
        if(digits == infinity || digits == notANumber) {
            if(digits == notANumber)
                return Double.NaN;
            else
                return (isNegative?Double.NEGATIVE_INFINITY:Double.POSITIVE_INFINITY);
        }
        else {
            if (mustSetRoundDir) {
                roundDir = 0;
            }
            /*
             * convert the lead kDigits to a long integer.
             */
            // (special performance hack: start to do it using int)
            int iValue = (int)digits[0]-(int)'0';
            int iDigits = Math.min( kDigits, intDecimalDigits );
            for ( int i=1; i < iDigits; i++ ){
                iValue = iValue*10 + (int)digits[i]-(int)'0';
            }
            lValue = (long)iValue;
            for ( int i=iDigits; i < kDigits; i++ ){
                lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
            }
            dValue = (double)lValue;
            int exp = decExponent-kDigits;
            /*
             * lValue now contains a long integer with the value of
             * the first kDigits digits of the number.
             * dValue contains the (double) of the same.
             */

            if ( nDigits <= maxDecimalDigits ){
                /*
                 * possibly an easy case.
                 * We know that the digits can be represented
                 * exactly. And if the exponent isn't too outrageous,
                 * the whole thing can be done with one operation,
                 * thus one rounding error.
                 * Note that all our constructors trim all leading and
                 * trailing zeros, so simple values (including zero)
                 * will always end up here
                 */
                if (exp == 0 || dValue == 0.0)
                    return (isNegative)? -dValue : dValue; // small floating integer
                else if ( exp >= 0 ){
                    if ( exp <= maxSmallTen ){
                        /*
                         * Can get the answer with one operation,
                         * thus one roundoff.
                         */
                        rValue = dValue * small10pow[exp];
                        if ( mustSetRoundDir ){
                            tValue = rValue / small10pow[exp];
                            roundDir = ( tValue ==  dValue ) ? 0
                                :( tValue < dValue ) ? 1
                                : -1;
                        }
                        return (isNegative)? -rValue : rValue;
                    }
                    int slop = maxDecimalDigits - kDigits;
                    if ( exp <= maxSmallTen+slop ){
                        /*
                         * We can multiply dValue by 10^(slop)
                         * and it is still "small" and exact.
                         * Then we can multiply by 10^(exp-slop)
                         * with one rounding.
                         */
                        dValue *= small10pow[slop];
                        rValue = dValue * small10pow[exp-slop];

                        if ( mustSetRoundDir ){
                            tValue = rValue / small10pow[exp-slop];
                            roundDir = ( tValue ==  dValue ) ? 0
                                :( tValue < dValue ) ? 1
                                : -1;
                        }
                        return (isNegative)? -rValue : rValue;
                    }
                    /*
                     * Else we have a hard case with a positive exp.
                     */
                } else {
                    if ( exp >= -maxSmallTen ){
                        /*
                         * Can get the answer in one division.
                         */
                        rValue = dValue / small10pow[-exp];
                        tValue = rValue * small10pow[-exp];
                        if ( mustSetRoundDir ){
                            roundDir = ( tValue ==  dValue ) ? 0
                                :( tValue < dValue ) ? 1
                                : -1;
                        }
                        return (isNegative)? -rValue : rValue;
                    }
                    /*
                     * Else we have a hard case with a negative exp.
                     */
                }
            }

            /*
             * Harder cases:
             * The sum of digits plus exponent is greater than
             * what we think we can do with one error.
             *
             * Start by approximating the right answer by,
             * naively, scaling by powers of 10.
             */
            if ( exp > 0 ){
                if ( decExponent > maxDecimalExponent+1 ){
                    /*
                     * Lets face it. This is going to be
                     * Infinity. Cut to the chase.
                     */
                    return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
                }
                if ( (exp&15) != 0 ){
                    dValue *= small10pow[exp&15];
                }
                if ( (exp>>=4) != 0 ){
                    int j;
                    for( j = 0; exp > 1; j++, exp>>=1 ){
                        if ( (exp&1)!=0)
                            dValue *= big10pow[j];
                    }
                    /*
                     * The reason for the weird exp > 1 condition
                     * in the above loop was so that the last multiply
                     * would get unrolled. We handle it here.
                     * It could overflow.
                     */
                    double t = dValue * big10pow[j];
                    if ( Double.isInfinite( t ) ){
                        /*
                         * It did overflow.
                         * Look more closely at the result.
                         * If the exponent is just one too large,
                         * then use the maximum finite as our estimate
                         * value. Else call the result infinity
                         * and punt it.
                         * ( I presume this could happen because
                         * rounding forces the result here to be
                         * an ULP or two larger than
                         * Double.MAX_VALUE ).
                         */
                        t = dValue / 2.0;
                        t *= big10pow[j];
                        if ( Double.isInfinite( t ) ){
                            return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
                        }
                        t = Double.MAX_VALUE;
                    }
                    dValue = t;
                }
            } else if ( exp < 0 ){
                exp = -exp;
                if ( decExponent < minDecimalExponent-1 ){
                    /*
                     * Lets face it. This is going to be
                     * zero. Cut to the chase.
                     */
                    return (isNegative)? -0.0 : 0.0;
                }
                if ( (exp&15) != 0 ){
                    dValue /= small10pow[exp&15];
                }
                if ( (exp>>=4) != 0 ){
                    int j;
                    for( j = 0; exp > 1; j++, exp>>=1 ){
                        if ( (exp&1)!=0)
                            dValue *= tiny10pow[j];
                    }
                    /*
                     * The reason for the weird exp > 1 condition
                     * in the above loop was so that the last multiply
                     * would get unrolled. We handle it here.
                     * It could underflow.
                     */
                    double t = dValue * tiny10pow[j];
                    if ( t == 0.0 ){
                        /*
                         * It did underflow.
                         * Look more closely at the result.
                         * If the exponent is just one too small,
                         * then use the minimum finite as our estimate
                         * value. Else call the result 0.0
                         * and punt it.
                         * ( I presume this could happen because
                         * rounding forces the result here to be
                         * an ULP or two less than
                         * Double.MIN_VALUE ).
                         */
                        t = dValue * 2.0;
                        t *= tiny10pow[j];
                        if ( t == 0.0 ){
                            return (isNegative)? -0.0 : 0.0;
                        }
                        t = Double.MIN_VALUE;
                    }
                    dValue = t;
                }
            }

            /*
             * dValue is now approximately the result.
             * The hard part is adjusting it, by comparison
             * with OldFDBigIntForTest arithmetic.
             * Formulate the EXACT big-number result as
             * bigD0 * 10^exp
             */
            OldFDBigIntForTest bigD0 = new OldFDBigIntForTest( lValue, digits, kDigits, nDigits );
            exp   = decExponent - nDigits;

            correctionLoop:
            while(true){
                /* AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES
                 * bigIntExp and bigIntNBits
                 */
                OldFDBigIntForTest bigB = doubleToBigInt( dValue );

                /*
                 * Scale bigD, bigB appropriately for
                 * big-integer operations.
                 * Naively, we multiply by powers of ten
                 * and powers of two. What we actually do
                 * is keep track of the powers of 5 and
                 * powers of 2 we would use, then factor out
                 * common divisors before doing the work.
                 */
                int B2, B5; // powers of 2, 5 in bigB
                int     D2, D5; // powers of 2, 5 in bigD
                int Ulp2;   // powers of 2 in halfUlp.
                if ( exp >= 0 ){
                    B2 = B5 = 0;
                    D2 = D5 = exp;
                } else {
                    B2 = B5 = -exp;
                    D2 = D5 = 0;
                }
                if ( bigIntExp >= 0 ){
                    B2 += bigIntExp;
                } else {
                    D2 -= bigIntExp;
                }
                Ulp2 = B2;
                // shift bigB and bigD left by a number s. t.
                // halfUlp is still an integer.
                int hulpbias;
                if ( bigIntExp+bigIntNBits <= -expBias+1 ){
                    // This is going to be a denormalized number
                    // (if not actually zero).
                    // half an ULP is at 2^-(expBias+expShift+1)
                    hulpbias = bigIntExp+ expBias + expShift;
                } else {
                    hulpbias = expShift + 2 - bigIntNBits;
                }
                B2 += hulpbias;
                D2 += hulpbias;
                // if there are common factors of 2, we might just as well
                // factor them out, as they add nothing useful.
                int common2 = Math.min( B2, Math.min( D2, Ulp2 ) );
                B2 -= common2;
                D2 -= common2;
                Ulp2 -= common2;
                // do multiplications by powers of 5 and 2
                bigB = multPow52( bigB, B5, B2 );
                OldFDBigIntForTest bigD = multPow52( new OldFDBigIntForTest( bigD0 ), D5, D2 );
                //
                // to recap:
                // bigB is the scaled-big-int version of our floating-point
                // candidate.
                // bigD is the scaled-big-int version of the exact value
                // as we understand it.
                // halfUlp is 1/2 an ulp of bigB, except for special cases
                // of exact powers of 2
                //
                // the plan is to compare bigB with bigD, and if the difference
                // is less than halfUlp, then we're satisfied. Otherwise,
                // use the ratio of difference to halfUlp to calculate a fudge
                // factor to add to the floating value, then go 'round again.
                //
                OldFDBigIntForTest diff;
                int cmpResult;
                boolean overvalue;
                if ( (cmpResult = bigB.cmp( bigD ) ) > 0 ){
                    overvalue = true; // our candidate is too big.
                    diff = bigB.sub( bigD );
                    if ( (bigIntNBits == 1) && (bigIntExp > -expBias+1) ){
                        // candidate is a normalized exact power of 2 and
                        // is too big. We will be subtracting.
                        // For our purposes, ulp is the ulp of the
                        // next smaller range.
                        Ulp2 -= 1;
                        if ( Ulp2 < 0 ){
                            // rats. Cannot de-scale ulp this far.
                            // must scale diff in other direction.
                            Ulp2 = 0;
                            diff.lshiftMe( 1 );
                        }
                    }
                } else if ( cmpResult < 0 ){
                    overvalue = false; // our candidate is too small.
                    diff = bigD.sub( bigB );
                } else {
                    // the candidate is exactly right!
                    // this happens with surprising frequency
                    break correctionLoop;
                }
                OldFDBigIntForTest halfUlp = constructPow52( B5, Ulp2 );
                if ( (cmpResult = diff.cmp( halfUlp ) ) < 0 ){
                    // difference is small.
                    // this is close enough
                    if (mustSetRoundDir) {
                        roundDir = overvalue ? -1 : 1;
                    }
                    break correctionLoop;
                } else if ( cmpResult == 0 ){
                    // difference is exactly half an ULP
                    // round to some other value maybe, then finish
                    dValue += 0.5*ulp( dValue, overvalue );
                    // should check for bigIntNBits == 1 here??
                    if (mustSetRoundDir) {
                        roundDir = overvalue ? -1 : 1;
                    }
                    break correctionLoop;
                } else {
                    // difference is non-trivial.
                    // could scale addend by ratio of difference to
                    // halfUlp here, if we bothered to compute that difference.
                    // Most of the time ( I hope ) it is about 1 anyway.
                    dValue += ulp( dValue, overvalue );
                    if ( dValue == 0.0 || dValue == Double.POSITIVE_INFINITY )
                        break correctionLoop; // oops. Fell off end of range.
                    continue; // try again.
                }

            }
            return (isNegative)? -dValue : dValue;
        }
    }

    /*
     * Take a FloatingDecimal, which we presumably just scanned in,
     * and find out what its value is, as a float.
     * This is distinct from doubleValue() to avoid the extremely
     * unlikely case of a double rounding error, wherein the conversion
     * to double has one rounding error, and the conversion of that double
     * to a float has another rounding error, IN THE WRONG DIRECTION,
     * ( because of the preference to a zero low-order bit ).
     */

    public strictfp float floatValue(){
        int     kDigits = Math.min( nDigits, singleMaxDecimalDigits+1 );
        int     iValue;
        float   fValue;

        // First, check for NaN and Infinity values
        if(digits == infinity || digits == notANumber) {
            if(digits == notANumber)
                return Float.NaN;
            else
                return (isNegative?Float.NEGATIVE_INFINITY:Float.POSITIVE_INFINITY);
        }
        else {
            /*
             * convert the lead kDigits to an integer.
             */
            iValue = (int)digits[0]-(int)'0';
            for ( int i=1; i < kDigits; i++ ){
                iValue = iValue*10 + (int)digits[i]-(int)'0';
            }
            fValue = (float)iValue;
            int exp = decExponent-kDigits;
            /*
             * iValue now contains an integer with the value of
             * the first kDigits digits of the number.
             * fValue contains the (float) of the same.
             */

            if ( nDigits <= singleMaxDecimalDigits ){
                /*
                 * possibly an easy case.
                 * We know that the digits can be represented
                 * exactly. And if the exponent isn't too outrageous,
                 * the whole thing can be done with one operation,
                 * thus one rounding error.
                 * Note that all our constructors trim all leading and
                 * trailing zeros, so simple values (including zero)
                 * will always end up here.
                 */
                if (exp == 0 || fValue == 0.0f)
                    return (isNegative)? -fValue : fValue; // small floating integer
                else if ( exp >= 0 ){
                    if ( exp <= singleMaxSmallTen ){
                        /*
                         * Can get the answer with one operation,
                         * thus one roundoff.
                         */
                        fValue *= singleSmall10pow[exp];
                        return (isNegative)? -fValue : fValue;
                    }
                    int slop = singleMaxDecimalDigits - kDigits;
                    if ( exp <= singleMaxSmallTen+slop ){
                        /*
                         * We can multiply dValue by 10^(slop)
                         * and it is still "small" and exact.
                         * Then we can multiply by 10^(exp-slop)
                         * with one rounding.
                         */
                        fValue *= singleSmall10pow[slop];
                        fValue *= singleSmall10pow[exp-slop];
                        return (isNegative)? -fValue : fValue;
                    }
                    /*
                     * Else we have a hard case with a positive exp.
                     */
                } else {
                    if ( exp >= -singleMaxSmallTen ){
                        /*
                         * Can get the answer in one division.
                         */
                        fValue /= singleSmall10pow[-exp];
                        return (isNegative)? -fValue : fValue;
                    }
                    /*
                     * Else we have a hard case with a negative exp.
                     */
                }
            } else if ( (decExponent >= nDigits) && (nDigits+decExponent <= maxDecimalDigits) ){
                /*
                 * In double-precision, this is an exact floating integer.
                 * So we can compute to double, then shorten to float
                 * with one round, and get the right answer.
                 *
                 * First, finish accumulating digits.
                 * Then convert that integer to a double, multiply
                 * by the appropriate power of ten, and convert to float.
                 */
                long lValue = (long)iValue;
                for ( int i=kDigits; i < nDigits; i++ ){
                    lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
                }
                double dValue = (double)lValue;
                exp = decExponent-nDigits;
                dValue *= small10pow[exp];
                fValue = (float)dValue;
                return (isNegative)? -fValue : fValue;

            }
            /*
             * Harder cases:
             * The sum of digits plus exponent is greater than
             * what we think we can do with one error.
             *
             * Start by weeding out obviously out-of-range
             * results, then convert to double and go to
             * common hard-case code.
             */
            if ( decExponent > singleMaxDecimalExponent+1 ){
                /*
                 * Lets face it. This is going to be
                 * Infinity. Cut to the chase.
                 */
                return (isNegative)? Float.NEGATIVE_INFINITY : Float.POSITIVE_INFINITY;
            } else if ( decExponent < singleMinDecimalExponent-1 ){
                /*
                 * Lets face it. This is going to be
                 * zero. Cut to the chase.
                 */
                return (isNegative)? -0.0f : 0.0f;
            }

            /*
             * Here, we do 'way too much work, but throwing away
             * our partial results, and going and doing the whole
             * thing as double, then throwing away half the bits that computes
             * when we convert back to float.
             *
             * The alternative is to reproduce the whole multiple-precision
             * algorithm for float precision, or to try to parameterize it
             * for common usage. The former will take about 400 lines of code,
             * and the latter I tried without success. Thus the semi-hack
             * answer here.
             */
            mustSetRoundDir = !fromHex;
            double dValue = doubleValue();
            return stickyRound( dValue );
        }
    }


    /*
     * All the positive powers of 10 that can be
     * represented exactly in double/float.
     */
    private static final double small10pow[] = {
        1.0e0,
        1.0e1, 1.0e2, 1.0e3, 1.0e4, 1.0e5,
        1.0e6, 1.0e7, 1.0e8, 1.0e9, 1.0e10,
        1.0e11, 1.0e12, 1.0e13, 1.0e14, 1.0e15,
        1.0e16, 1.0e17, 1.0e18, 1.0e19, 1.0e20,
        1.0e21, 1.0e22
    };

    private static final float singleSmall10pow[] = {
        1.0e0f,
        1.0e1f, 1.0e2f, 1.0e3f, 1.0e4f, 1.0e5f,
        1.0e6f, 1.0e7f, 1.0e8f, 1.0e9f, 1.0e10f
    };

    private static final double big10pow[] = {
        1e16, 1e32, 1e64, 1e128, 1e256 };
    private static final double tiny10pow[] = {
        1e-16, 1e-32, 1e-64, 1e-128, 1e-256 };

    private static final int maxSmallTen = small10pow.length-1;
    private static final int singleMaxSmallTen = singleSmall10pow.length-1;

    private static final int small5pow[] = {
        1,
        5,
        5*5,
        5*5*5,
        5*5*5*5,
        5*5*5*5*5,
        5*5*5*5*5*5,
        5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5*5*5*5*5,
        5*5*5*5*5*5*5*5*5*5*5*5*5
    };


    private static final long long5pow[] = {
        1L,
        5L,
        5L*5,
        5L*5*5,
        5L*5*5*5,
        5L*5*5*5*5,
        5L*5*5*5*5*5,
        5L*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
        5L*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5*5,
    };

    // approximately ceil( log2( long5pow[i] ) )
    private static final int n5bits[] = {
        0,
        3,
        5,
        7,
        10,
        12,
        14,
        17,
        19,
        21,
        24,
        26,
        28,
        31,
        33,
        35,
        38,
        40,
        42,
        45,
        47,
        49,
        52,
        54,
        56,
        59,
        61,
    };

    private static final char infinity[] = { 'I', 'n', 'f', 'i', 'n', 'i', 't', 'y' };
    private static final char notANumber[] = { 'N', 'a', 'N' };
    private static final char zero[] = { '0', '0', '0', '0', '0', '0', '0', '0' };


    /*
     * Grammar is compatible with hexadecimal floating-point constants
     * described in section 6.4.4.2 of the C99 specification.
     */
    private static Pattern hexFloatPattern = null;
    private static synchronized Pattern getHexFloatPattern() {
        if (hexFloatPattern == null) {
           hexFloatPattern = Pattern.compile(
                   //1           234                   56                7                   8      9
                    "([-+])?0[xX](((\\p{XDigit}+)\\.?)|((\\p{XDigit}*)\\.(\\p{XDigit}+)))[pP]([-+])?(\\p{Digit}+)[fFdD]?"
                    );
        }
        return hexFloatPattern;
    }

    /*
     * Convert string s to a suitable floating decimal; uses the
     * double constructor and set the roundDir variable appropriately
     * in case the value is later converted to a float.
     */
   static OldFloatingDecimalForTest parseHexString(String s) {
        // Verify string is a member of the hexadecimal floating-point
        // string language.
        Matcher m = getHexFloatPattern().matcher(s);
        boolean validInput = m.matches();

        if (!validInput) {
            // Input does not match pattern
            throw new NumberFormatException("For input string: \"" + s + "\"");
        } else { // validInput
            /*
             * We must isolate the sign, significand, and exponent
             * fields.  The sign value is straightforward.  Since
             * floating-point numbers are stored with a normalized
             * representation, the significand and exponent are
             * interrelated.
             *
             * After extracting the sign, we normalized the
             * significand as a hexadecimal value, calculating an
             * exponent adjust for any shifts made during
             * normalization.  If the significand is zero, the
             * exponent doesn't need to be examined since the output
             * will be zero.
             *
             * Next the exponent in the input string is extracted.
             * Afterwards, the significand is normalized as a *binary*
             * value and the input value's normalized exponent can be
             * computed.  The significand bits are copied into a
             * double significand; if the string has more logical bits
             * than can fit in a double, the extra bits affect the
             * round and sticky bits which are used to round the final
             * value.
             */

            //  Extract significand sign
            String group1 = m.group(1);
            double sign = (( group1 == null ) || group1.equals("+"))? 1.0 : -1.0;


            //  Extract Significand magnitude
            /*
             * Based on the form of the significand, calculate how the
             * binary exponent needs to be adjusted to create a
             * normalized *hexadecimal* floating-point number; that
             * is, a number where there is one nonzero hex digit to
             * the left of the (hexa)decimal point.  Since we are
             * adjusting a binary, not hexadecimal exponent, the
             * exponent is adjusted by a multiple of 4.
             *
             * There are a number of significand scenarios to consider;
             * letters are used in indicate nonzero digits:
             *
             * 1. 000xxxx       =>      x.xxx   normalized
             *    increase exponent by (number of x's - 1)*4
             *
             * 2. 000xxx.yyyy =>        x.xxyyyy        normalized
             *    increase exponent by (number of x's - 1)*4
             *
             * 3. .000yyy  =>   y.yy    normalized
             *    decrease exponent by (number of zeros + 1)*4
             *
             * 4. 000.00000yyy => y.yy normalized
             *    decrease exponent by (number of zeros to right of point + 1)*4
             *
             * If the significand is exactly zero, return a properly
             * signed zero.
             */

            String significandString =null;
            int signifLength = 0;
            int exponentAdjust = 0;
            {
                int leftDigits  = 0; // number of meaningful digits to
                                     // left of "decimal" point
                                     // (leading zeros stripped)
                int rightDigits = 0; // number of digits to right of
                                     // "decimal" point; leading zeros
                                     // must always be accounted for
                /*
                 * The significand is made up of either
                 *
                 * 1. group 4 entirely (integer portion only)
                 *
                 * OR
                 *
                 * 2. the fractional portion from group 7 plus any
                 * (optional) integer portions from group 6.
                 */
                String group4;
                if( (group4 = m.group(4)) != null) {  // Integer-only significand
                    // Leading zeros never matter on the integer portion
                    significandString = stripLeadingZeros(group4);
                    leftDigits = significandString.length();
                }
                else {
                    // Group 6 is the optional integer; leading zeros
                    // never matter on the integer portion
                    String group6 = stripLeadingZeros(m.group(6));
                    leftDigits = group6.length();

                    // fraction
                    String group7 = m.group(7);
                    rightDigits = group7.length();

                    // Turn "integer.fraction" into "integer"+"fraction"
                    significandString =
                        ((group6 == null)?"":group6) + // is the null
                        // check necessary?
                        group7;
                }

                significandString = stripLeadingZeros(significandString);
                signifLength  = significandString.length();

                /*
                 * Adjust exponent as described above
                 */
                if (leftDigits >= 1) {  // Cases 1 and 2
                    exponentAdjust = 4*(leftDigits - 1);
                } else {                // Cases 3 and 4
                    exponentAdjust = -4*( rightDigits - signifLength + 1);
                }

                // If the significand is zero, the exponent doesn't
                // matter; return a properly signed zero.

                if (signifLength == 0) { // Only zeros in input
                    return new OldFloatingDecimalForTest(sign * 0.0);
                }
            }

            //  Extract Exponent
            /*
             * Use an int to read in the exponent value; this should
             * provide more than sufficient range for non-contrived
             * inputs.  If reading the exponent in as an int does
             * overflow, examine the sign of the exponent and
             * significand to determine what to do.
             */
            String group8 = m.group(8);
            boolean positiveExponent = ( group8 == null ) || group8.equals("+");
            long unsignedRawExponent;
            try {
                unsignedRawExponent = Integer.parseInt(m.group(9));
            }
            catch (NumberFormatException e) {
                // At this point, we know the exponent is
                // syntactically well-formed as a sequence of
                // digits.  Therefore, if an NumberFormatException
                // is thrown, it must be due to overflowing int's
                // range.  Also, at this point, we have already
                // checked for a zero significand.  Thus the signs
                // of the exponent and significand determine the
                // final result:
                //
                //                      significand
                //                      +               -
                // exponent     +       +infinity       -infinity
                //              -       +0.0            -0.0
                return new OldFloatingDecimalForTest(sign * (positiveExponent ?
                                                   Double.POSITIVE_INFINITY : 0.0));
            }

            long rawExponent =
                (positiveExponent ? 1L : -1L) * // exponent sign
                unsignedRawExponent;            // exponent magnitude

            // Calculate partially adjusted exponent
            long exponent = rawExponent + exponentAdjust ;

            // Starting copying non-zero bits into proper position in
            // a long; copy explicit bit too; this will be masked
            // later for normal values.

            boolean round = false;
            boolean sticky = false;
            int bitsCopied=0;
            int nextShift=0;
            long significand=0L;
            // First iteration is different, since we only copy
            // from the leading significand bit; one more exponent
            // adjust will be needed...

            // IMPORTANT: make leadingDigit a long to avoid
            // surprising shift semantics!
            long leadingDigit = getHexDigit(significandString, 0);

            /*
             * Left shift the leading digit (53 - (bit position of
             * leading 1 in digit)); this sets the top bit of the
             * significand to 1.  The nextShift value is adjusted
             * to take into account the number of bit positions of
             * the leadingDigit actually used.  Finally, the
             * exponent is adjusted to normalize the significand
             * as a binary value, not just a hex value.
             */
            if (leadingDigit == 1) {
                significand |= leadingDigit << 52;
                nextShift = 52 - 4;
                /* exponent += 0 */     }
            else if (leadingDigit <= 3) { // [2, 3]
                significand |= leadingDigit << 51;
                nextShift = 52 - 5;
                exponent += 1;
            }
            else if (leadingDigit <= 7) { // [4, 7]
                significand |= leadingDigit << 50;
                nextShift = 52 - 6;
                exponent += 2;
            }
            else if (leadingDigit <= 15) { // [8, f]
                significand |= leadingDigit << 49;
                nextShift = 52 - 7;
                exponent += 3;
            } else {
                throw new AssertionError("Result from digit conversion too large!");
            }
            // The preceding if-else could be replaced by a single
            // code block based on the high-order bit set in
            // leadingDigit.  Given leadingOnePosition,

            // significand |= leadingDigit << (SIGNIFICAND_WIDTH - leadingOnePosition);
            // nextShift = 52 - (3 + leadingOnePosition);
            // exponent += (leadingOnePosition-1);


            /*
             * Now the exponent variable is equal to the normalized
             * binary exponent.  Code below will make representation
             * adjustments if the exponent is incremented after
             * rounding (includes overflows to infinity) or if the
             * result is subnormal.
             */

            // Copy digit into significand until the significand can't
            // hold another full hex digit or there are no more input
            // hex digits.
            int i = 0;
            for(i = 1;
                i < signifLength && nextShift >= 0;
                i++) {
                long currentDigit = getHexDigit(significandString, i);
                significand |= (currentDigit << nextShift);
                nextShift-=4;
            }

            // After the above loop, the bulk of the string is copied.
            // Now, we must copy any partial hex digits into the
            // significand AND compute the round bit and start computing
            // sticky bit.

            if ( i < signifLength ) { // at least one hex input digit exists
                long currentDigit = getHexDigit(significandString, i);

                // from nextShift, figure out how many bits need
                // to be copied, if any
                switch(nextShift) { // must be negative
                case -1:
                    // three bits need to be copied in; can
                    // set round bit
                    significand |= ((currentDigit & 0xEL) >> 1);
                    round = (currentDigit & 0x1L)  != 0L;
                    break;

                case -2:
                    // two bits need to be copied in; can
                    // set round and start sticky
                    significand |= ((currentDigit & 0xCL) >> 2);
                    round = (currentDigit &0x2L)  != 0L;
                    sticky = (currentDigit & 0x1L) != 0;
                    break;

                case -3:
                    // one bit needs to be copied in
                    significand |= ((currentDigit & 0x8L)>>3);
                    // Now set round and start sticky, if possible
                    round = (currentDigit &0x4L)  != 0L;
                    sticky = (currentDigit & 0x3L) != 0;
                    break;

                case -4:
                    // all bits copied into significand; set
                    // round and start sticky
                    round = ((currentDigit & 0x8L) != 0);  // is top bit set?
                    // nonzeros in three low order bits?
                    sticky = (currentDigit & 0x7L) != 0;
                    break;

                default:
                    throw new AssertionError("Unexpected shift distance remainder.");
                    // break;
                }

                // Round is set; sticky might be set.

                // For the sticky bit, it suffices to check the
                // current digit and test for any nonzero digits in
                // the remaining unprocessed input.
                i++;
                while(i < signifLength && !sticky) {
                    currentDigit =  getHexDigit(significandString,i);
                    sticky = sticky || (currentDigit != 0);
                    i++;
                }

            }
            // else all of string was seen, round and sticky are
            // correct as false.


            // Check for overflow and update exponent accordingly.

            if (exponent > Double.MAX_EXPONENT) {         // Infinite result
                // overflow to properly signed infinity
                return new OldFloatingDecimalForTest(sign * Double.POSITIVE_INFINITY);
            } else {  // Finite return value
                if (exponent <= Double.MAX_EXPONENT && // (Usually) normal result
                    exponent >= Double.MIN_EXPONENT) {

                    // The result returned in this block cannot be a
                    // zero or subnormal; however after the
                    // significand is adjusted from rounding, we could
                    // still overflow in infinity.

                    // AND exponent bits into significand; if the
                    // significand is incremented and overflows from
                    // rounding, this combination will update the
                    // exponent correctly, even in the case of
                    // Double.MAX_VALUE overflowing to infinity.

                    significand = (( (exponent +
                                     (long)DoubleConsts.EXP_BIAS) <<
                                     (DoubleConsts.SIGNIFICAND_WIDTH-1))
                                   & DoubleConsts.EXP_BIT_MASK) |
                        (DoubleConsts.SIGNIF_BIT_MASK & significand);

                }  else  {  // Subnormal or zero
                    // (exponent < Double.MIN_EXPONENT)

                    if (exponent < (DoubleConsts.MIN_SUB_EXPONENT -1 )) {
                        // No way to round back to nonzero value
                        // regardless of significand if the exponent is
                        // less than -1075.
                        return new OldFloatingDecimalForTest(sign * 0.0);
                    } else { //  -1075 <= exponent <= MIN_EXPONENT -1 = -1023
                        /*
                         * Find bit position to round to; recompute
                         * round and sticky bits, and shift
                         * significand right appropriately.
                         */

                        sticky = sticky || round;
                        round = false;

                        // Number of bits of significand to preserve is
                        // exponent - abs_min_exp +1
                        // check:
                        // -1075 +1074 + 1 = 0
                        // -1023 +1074 + 1 = 52

                        int bitsDiscarded = 53 -
                            ((int)exponent - DoubleConsts.MIN_SUB_EXPONENT + 1);
                        assert bitsDiscarded >= 1 && bitsDiscarded <= 53;

                        // What to do here:
                        // First, isolate the new round bit
                        round = (significand & (1L << (bitsDiscarded -1))) != 0L;
                        if (bitsDiscarded > 1) {
                            // create mask to update sticky bits; low
                            // order bitsDiscarded bits should be 1
                            long mask = ~((~0L) << (bitsDiscarded -1));
                            sticky = sticky || ((significand & mask) != 0L ) ;
                        }

                        // Now, discard the bits
                        significand = significand >> bitsDiscarded;

                        significand = (( ((long)(Double.MIN_EXPONENT -1) + // subnorm exp.
                                          (long)DoubleConsts.EXP_BIAS) <<
                                         (DoubleConsts.SIGNIFICAND_WIDTH-1))
                                       & DoubleConsts.EXP_BIT_MASK) |
                            (DoubleConsts.SIGNIF_BIT_MASK & significand);
                    }
                }

                // The significand variable now contains the currently
                // appropriate exponent bits too.

                /*
                 * Determine if significand should be incremented;
                 * making this determination depends on the least
                 * significant bit and the round and sticky bits.
                 *
                 * Round to nearest even rounding table, adapted from
                 * table 4.7 in "Computer Arithmetic" by IsraelKoren.
                 * The digit to the left of the "decimal" point is the
                 * least significant bit, the digits to the right of
                 * the point are the round and sticky bits
                 *
                 * Number       Round(x)
                 * x0.00        x0.
                 * x0.01        x0.
                 * x0.10        x0.
                 * x0.11        x1. = x0. +1
                 * x1.00        x1.
                 * x1.01        x1.
                 * x1.10        x1. + 1
                 * x1.11        x1. + 1
                 */
                boolean incremented = false;
                boolean leastZero  = ((significand & 1L) == 0L);
                if( (  leastZero  && round && sticky ) ||
                    ((!leastZero) && round )) {
                    incremented = true;
                    significand++;
                }

                OldFloatingDecimalForTest fd = new OldFloatingDecimalForTest(Math.copySign(
                                                              Double.longBitsToDouble(significand),
                                                              sign));

                /*
                 * Set roundingDir variable field of fd properly so
                 * that the input string can be properly rounded to a
                 * float value.  There are two cases to consider:
                 *
                 * 1. rounding to double discards sticky bit
                 * information that would change the result of a float
                 * rounding (near halfway case between two floats)
                 *
                 * 2. rounding to double rounds up when rounding up
                 * would not occur when rounding to float.
                 *
                 * For former case only needs to be considered when
                 * the bits rounded away when casting to float are all
                 * zero; otherwise, float round bit is properly set
                 * and sticky will already be true.
                 *
                 * The lower exponent bound for the code below is the
                 * minimum (normalized) subnormal exponent - 1 since a
                 * value with that exponent can round up to the
                 * minimum subnormal value and the sticky bit
                 * information must be preserved (i.e. case 1).
                 */
                if ((exponent >= FloatConsts.MIN_SUB_EXPONENT-1) &&
                    (exponent <= Float.MAX_EXPONENT ) ){
                    // Outside above exponent range, the float value
                    // will be zero or infinity.

                    /*
                     * If the low-order 28 bits of a rounded double
                     * significand are 0, the double could be a
                     * half-way case for a rounding to float.  If the
                     * double value is a half-way case, the double
                     * significand may have to be modified to round
                     * the the right float value (see the stickyRound
                     * method).  If the rounding to double has lost
                     * what would be float sticky bit information, the
                     * double significand must be incremented.  If the
                     * double value's significand was itself
                     * incremented, the float value may end up too
                     * large so the increment should be undone.
                     */
                    if ((significand & 0xfffffffL) ==  0x0L) {
                        // For negative values, the sign of the
                        // roundDir is the same as for positive values
                        // since adding 1 increasing the significand's
                        // magnitude and subtracting 1 decreases the
                        // significand's magnitude.  If neither round
                        // nor sticky is true, the double value is
                        // exact and no adjustment is required for a
                        // proper float rounding.
                        if( round || sticky) {
                            if (leastZero) { // prerounding lsb is 0
                                // If round and sticky were both true,
                                // and the least significant
                                // significand bit were 0, the rounded
                                // significand would not have its
                                // low-order bits be zero.  Therefore,
                                // we only need to adjust the
                                // significand if round XOR sticky is
                                // true.
                                if (round ^ sticky) {
                                    fd.roundDir =  1;
                                }
                            }
                            else { // prerounding lsb is 1
                                // If the prerounding lsb is 1 and the
                                // resulting significand has its
                                // low-order bits zero, the significand
                                // was incremented.  Here, we undo the
                                // increment, which will ensure the
                                // right guard and sticky bits for the
                                // float rounding.
                                if (round)
                                    fd.roundDir =  -1;
                            }
                        }
                    }
                }

                fd.fromHex = true;
                return fd;
            }
        }
    }

    /**
     * Return <code>s</code> with any leading zeros removed.
     */
    static String stripLeadingZeros(String s) {
        return  s.replaceFirst("^0+", "");
    }

    /**
     * Extract a hexadecimal digit from position <code>position</code>
     * of string <code>s</code>.
     */
    static int getHexDigit(String s, int position) {
        int value = Character.digit(s.charAt(position), 16);
        if (value <= -1 || value >= 16) {
            throw new AssertionError("Unexpected failure of digit conversion of " +
                                     s.charAt(position));
        }
        return value;
    }


}