1/* 2 * Copyright (c) 1980 Regents of the University of California. 3 * All rights reserved. 4 * 5 * Redistribution and use in source and binary forms are permitted 6 * provided that the above copyright notice and this paragraph are 7 * duplicated in all such forms and that any documentation, 8 * advertising materials, and other materials related to such 9 * distribution and use acknowledge that the software was developed 10 * by the University of California, Berkeley. The name of the 11 * University may not be used to endorse or promote products derived 12 * from this software without specific prior written permission. 13 * THIS SOFTWARE IS PROVIDED ``AS IS'' AND WITHOUT ANY EXPRESS OR 14 * IMPLIED WARRANTIES, INCLUDING, WITHOUT LIMITATION, THE IMPLIED 15 * WARRANTIES OF MERCHANTIBILITY AND FITNESS FOR A PARTICULAR PURPOSE. 16 */ 17 18#if defined(LIBC_SCCS) && !defined(lint) 19 .asciz "@(#)atof.s 5.5 (Berkeley) 06/27/88" 20#endif /* LIBC_SCCS and not lint */ 21 22#include "DEFS.h" 23 24/* 25 * atof: convert ascii to floating 26 * 27 * C usage: 28 * 29 * double atof (s) 30 * char *s; 31 * 32 * Register usage: 33 * 34 * r0-1: value being developed 35 * r2: first section: pointer to the next character 36 * second section: binary exponent 37 * r3: flags 38 * r4: first section: the current character 39 * second section: scratch 40 * r5: the decimal exponent 41 * r6-7: scratch 42 */ 43 .set msign,0 # mantissa has negative sign 44 .set esign,1 # exponent has negative sign 45 .set decpt,2 # decimal point encountered 46 47ENTRY(atof, R6|R7) 48/* 49 * Initialization 50 */ 51 clrl r3 # All flags start out false 52 movl 4(ap),r2 # Address the first character 53 clrl r5 # Clear starting exponent 54/* 55 * Skip leading white space 56 */ 57sk0: movzbl (r2)+,r4 # Fetch the next (first) character 58 cmpb $' ,r4 # Is it blank? 59 jeql sk0 # ...yes 60 cmpb r4,$8 # 8 is lowest of white-space group 61 jlss sk1 # Jump if char too low to be white space 62 cmpb r4,$13 # 13 is highest of white-space group 63 jleq sk0 # Jump if character is white space 64sk1: 65/* 66 * Check for a sign 67 */ 68 cmpb $'+,r4 # Positive sign? 69 jeql cs1 # ... yes 70 cmpb $'-,r4 # Negative sign? 71 jneq cs2 # ... no 72 bisb2 $1<msign,r3 # Indicate a negative mantissa 73cs1: movzbl (r2)+,r4 # Skip the character 74cs2: 75/* 76 * Accumulate digits, keeping track of the exponent 77 */ 78 clrq r0 # Clear the accumulator 79ad0: cmpb r4,$'0 # Do we have a digit? 80 jlss ad4 # ... no, too small 81 cmpb r4,$'9 82 jgtr ad4 # ... no, too large 83/* 84 * We got a digit. Accumulate it 85 */ 86 cmpl r1,$214748364 # Would this digit cause overflow? 87 jgeq ad1 # ... yes 88/* 89 * Multiply (r0,r1) by 10. This is done by developing 90 * (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits, 91 * and adding the two quadwords. 92 */ 93 ashq $1,r0,r6 # (r6,r7)=(r0,r1)*2 94 ashq $3,r0,r0 # (r0,r1)=(r0,r1)*8 95 addl2 r6,r0 # Add low halves 96 adwc r7,r1 # Add high halves 97/* 98 * Add in the digit 99 */ 100 subl2 $'0,r4 # Get the digit value 101 addl2 r4,r0 # Add it into the accumulator 102 adwc $0,r1 # Possible carry into high half 103 jbr ad2 # Join common code 104/* 105 * Here when the digit won't fit in the accumulator 106 */ 107ad1: incl r5 # Ignore the digit, bump exponent 108/* 109 * If we have seen a decimal point, decrease the exponent by 1 110 */ 111ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen 112 decl r5 # Decrease exponent 113ad3: 114/* 115 * Fetch the next character, back for more 116 */ 117 movzbl (r2)+,r4 # Fetch 118 jbr ad0 # Try again 119/* 120 * Not a digit. Could it be a decimal point? 121 */ 122ad4: cmpb r4,$'. # If it's not a decimal point, either it's 123 jneq ad5 # the end of the number or the start of 124 # the exponent. 125 jbcs $decpt,r3,ad3 # If it IS a decimal point, we record that 126 # we've seen one, and keep collecting 127 # digits if it is the first one. 128/* 129 * Check for an exponent 130 */ 131ad5: clrl r6 # Initialize the exponent accumulator 132 133 cmpb r4,$'e # We allow both lower case e 134 jeql ex1 # ... and ... 135 cmpb r4,$'E # upper-case E 136 jneq ex7 137/* 138 * Does the exponent have a sign? 139 */ 140ex1: movzbl (r2)+,r4 # Get next character 141 cmpb r4,$'+ # Positive sign? 142 jeql ex2 # ... yes ... 143 cmpb r4,$'- # Negative sign? 144 jneq ex3 # ... no ... 145 bisb2 $1<esign,r3 # Indicate exponent is negative 146ex2: movzbl (r2)+,r4 # Grab the next character 147/* 148 * Accumulate exponent digits in r6 149 */ 150ex3: cmpb r4,$'0 # A digit is within the range 151 jlss ex4 # '0' through 152 cmpb r4,$'9 # '9', 153 jgtr ex4 # inclusive. 154 cmpl r6,$214748364 # Exponent outrageously large already? 155 jgeq ex2 # ... yes 156 moval (r6)[r6],r6 # r6 *= 5 157 movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0' 158 jbr ex2 # Go 'round again 159ex4: 160/* 161 * Now get the final exponent and force it within a reasonable 162 * range so our scaling loops don't take forever for values 163 * that will ultimately cause overflow or underflow anyway. 164 * A tight check on over/underflow will be done by ldexp. 165 */ 166 jbc $esign,r3,ex5 # Jump if exponent not negative 167 mnegl r6,r6 # If sign, negate exponent 168ex5: addl2 r6,r5 # Add given exponent to calculated exponent 169 cmpl r5,$-100 # Absurdly small? 170 jgtr ex6 # ... no 171 movl $-100,r5 # ... yes, force within limit 172ex6: cmpl r5,$100 # Absurdly large? 173 jlss ex7 # ... no 174 movl $100,r5 # ... yes, force within bounds 175ex7: 176/* 177 * Our number has now been reduced to a mantissa and an exponent. 178 * The mantissa is a 63-bit positive binary integer in r0,r1, 179 * and the exponent is a signed power of 10 in r5. The msign 180 * bit in r3 will be on if the mantissa should ultimately be 181 * considered negative. 182 * 183 * We now have to convert it to a standard format floating point 184 * number. This will be done by accumulating a binary exponent 185 * in r2, as we progressively get r5 closer to zero. 186 * 187 * Don't bother scaling if the mantissa is zero 188 */ 189 movq r0,r0 # Mantissa zero? 190 jeql exit # ... yes 191 192 clrl r2 # Initialize binary exponent 193 tstl r5 # Which way to scale? 194 jleq sd0 # Scale down if decimal exponent <= 0 195/* 196 * Scale up by "multiplying" r0,r1 by 10 as many times as necessary, 197 * as follows: 198 * 199 * Step 1: Shift r0,r1 right as necessary to ensure that no 200 * overflow can occur when multiplying. 201 */ 202su0: cmpl r1,$429496729 # Compare high word to (2**31)/5 203 jlss su1 # Jump out if guaranteed safe 204 ashq $-1,r0,r0 # Else shift right one bit 205 incl r2 # bump exponent to compensate 206 jbr su0 # and go back to test again. 207/* 208 * Step 2: Multiply r0,r1 by 5, by appropriate shifting and 209 * double-precision addition 210 */ 211su1: ashq $2,r0,r6 # (r6,r7) := (r0,r1) * 4 212 addl2 r6,r0 # Add low-order halves 213 adwc r7,r1 # and high-order halves 214/* 215 * Step 3: Increment the binary exponent to take care of the final 216 * factor of 2, and go back if we still need to scale more. 217 */ 218 incl r2 # Increment the exponent 219 sobgtr r5,su0 # and back for more (maybe) 220 221 jbr cm0 # Merge to build final value 222 223/* 224 * Scale down. We must "divide" r0,r1 by 10 as many times 225 * as needed, as follows: 226 * 227 * Step 0: Right now, the condition codes reflect the state 228 * of r5. If it's zero, we are done. 229 */ 230sd0: jeql cm0 # If finished, build final number 231/* 232 * Step 1: Shift r0,r1 left until the high-order bit (not counting 233 * the sign bit) is nonzero, so that the division will preserve 234 * as much precision as possible. 235 */ 236 tstl r1 # Is the entire high-order half zero? 237 jneq sd2 # ...no, go shift one bit at a time 238 ashq $30,r0,r0 # ...yes, shift left 30, 239 subl2 $30,r2 # decrement the exponent to compensate, 240 # and now it's known to be safe to shift 241 # at least once more. 242sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and 243 decl r2 # decrement the exponent to compensate 244sd2: jbc $30,r1,sd1 # If the high-order bit is off, go shift 245/* 246 * Step 2: Divide the high-order part of (r0,r1) by 5, 247 * giving a quotient in r1 and a remainder in r7. 248 */ 249sd3: movl r1,r6 # Copy the high-order part 250 clrl r7 # Zero-extend to 64 bits 251 ediv $5,r6,r1,r7 # Divide (cannot overflow) 252/* 253 * Step 3: Divide the low-order part of (r0,r1) by 5, 254 * using the remainder from step 2 for rounding. 255 * Note that the result of this computation is unsigned, 256 * so we have to allow for the fact that an ordinary division 257 * by 5 could overflow. We make allowance by dividing by 10, 258 * multiplying the quotient by 2, and using the remainder 259 * to adjust the modified quotient. 260 */ 261 addl3 $2,r0,r6 # Dividend is low part of (r0,r1) plus 262 adwc $0,r7 # 2 for rounding plus 263 # (2**32) * previous remainder 264 ediv $10,r6,r0,r6 # r0 := quotient, r6 := remainder. 265 addl2 r0,r0 # Make r0 result of dividing by 5 266 cmpl r6,$5 # If remainder is 5 or greater, 267 jlss sd4 # increment the adjustted quotient. 268 incl r0 269/* 270 * Step 4: Increment the decimal exponent, decrement the binary 271 * exponent (to make the division by 5 into a division by 10), 272 * and back for another iteration. 273 */ 274sd4: decl r2 # Binary exponent 275 aoblss $0,r5,sd2 276/* 277 * We now have the following: 278 * 279 * r0: low-order half of a 64-bit integer 280 * r1: high-order half of the same 64-bit integer 281 * r2: a binary exponent 282 * 283 * Our final result is the integer represented by (r0,r1) 284 * multiplied by 2 to the power contained in r2. 285 * We will transform (r0,r1) into a floating-point value, 286 * set the sign appropriately, and let ldexp do the 287 * rest of the work. 288 * 289 * Step 1: if the high-order bit (excluding the sign) of 290 * the high-order half (r1) is 1, then we have 63 bits of 291 * fraction, too many to convert easily. However, we also 292 * know we won't need them all, so we will just throw the 293 * low-order bit away (and adjust the exponent appropriately). 294 */ 295cm0: jbc $30,r1,cm1 # jump if no adjustment needed 296 ashq $-1,r0,r0 # lose the low-order bit 297 incl r2 # increase the exponent to compensate 298/* 299 * Step 2: split the 62-bit number in (r0,r1) into two 300 * 31-bit positive quantities 301 */ 302cm1: ashq $1,r0,r0 # put the high-order bits in r1 303 # and a 0 in the bottom of r0 304 rotl $-1,r0,r0 # right-justify the bits in r0 305 # moving the 0 from the ashq 306 # into the sign bit. 307/* 308 * Step 3: convert both halves to floating point 309 */ 310 cvtld r0,r6 # low-order part in r6-r7 311 cvtld r1,r0 # high-order part in r0-r1 312/* 313 * Step 4: multiply the high order part by 2**31 and combine them 314 */ 315 muld2 two31,r0 # multiply 316 addd2 r6,r0 # combine 317/* 318 * Step 5: if appropriate, negate the floating value 319 */ 320 jbc $msign,r3,cm2 # Jump if mantissa not signed 321 mnegd r0,r0 # If negative, make it so 322/* 323 * Step 6: call ldexp to complete the job 324 */ 325cm2: pushl r2 # Put exponent in parameter list 326 movd r0,-(sp) # and also mantissa 327 calls $3,_ldexp # go combine them 328 329exit: 330 ret 331 332 .align 2 333two31: .word 0x5000 # 2 ** 31 334 .word 0 # (=2147483648) 335 .word 0 # in floating-point 336 .word 0 # (so atof doesn't have to convert it) 337