1/* 2 * Copyright (c) 1980 Regents of the University of California. 3 * All rights reserved. 4 * 5 * %sccs.include.redist.c% 6 */ 7 8#if defined(LIBC_SCCS) && !defined(lint) 9 .asciz "@(#)atof.s 5.6 (Berkeley) 06/01/90" 10#endif /* LIBC_SCCS and not lint */ 11 12#include "DEFS.h" 13 14/* 15 * atof: convert ascii to floating 16 * 17 * C usage: 18 * 19 * double atof (s) 20 * char *s; 21 * 22 * Register usage: 23 * 24 * r0-1: value being developed 25 * r2: first section: pointer to the next character 26 * second section: binary exponent 27 * r3: flags 28 * r4: first section: the current character 29 * second section: scratch 30 * r5: the decimal exponent 31 * r6-7: scratch 32 */ 33 .set msign,0 # mantissa has negative sign 34 .set esign,1 # exponent has negative sign 35 .set decpt,2 # decimal point encountered 36 37ENTRY(atof, R6|R7) 38/* 39 * Initialization 40 */ 41 clrl r3 # All flags start out false 42 movl 4(ap),r2 # Address the first character 43 clrl r5 # Clear starting exponent 44/* 45 * Skip leading white space 46 */ 47sk0: movzbl (r2)+,r4 # Fetch the next (first) character 48 cmpb $' ,r4 # Is it blank? 49 jeql sk0 # ...yes 50 cmpb r4,$8 # 8 is lowest of white-space group 51 jlss sk1 # Jump if char too low to be white space 52 cmpb r4,$13 # 13 is highest of white-space group 53 jleq sk0 # Jump if character is white space 54sk1: 55/* 56 * Check for a sign 57 */ 58 cmpb $'+,r4 # Positive sign? 59 jeql cs1 # ... yes 60 cmpb $'-,r4 # Negative sign? 61 jneq cs2 # ... no 62 bisb2 $1<msign,r3 # Indicate a negative mantissa 63cs1: movzbl (r2)+,r4 # Skip the character 64cs2: 65/* 66 * Accumulate digits, keeping track of the exponent 67 */ 68 clrq r0 # Clear the accumulator 69ad0: cmpb r4,$'0 # Do we have a digit? 70 jlss ad4 # ... no, too small 71 cmpb r4,$'9 72 jgtr ad4 # ... no, too large 73/* 74 * We got a digit. Accumulate it 75 */ 76 cmpl r1,$214748364 # Would this digit cause overflow? 77 jgeq ad1 # ... yes 78/* 79 * Multiply (r0,r1) by 10. This is done by developing 80 * (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits, 81 * and adding the two quadwords. 82 */ 83 ashq $1,r0,r6 # (r6,r7)=(r0,r1)*2 84 ashq $3,r0,r0 # (r0,r1)=(r0,r1)*8 85 addl2 r6,r0 # Add low halves 86 adwc r7,r1 # Add high halves 87/* 88 * Add in the digit 89 */ 90 subl2 $'0,r4 # Get the digit value 91 addl2 r4,r0 # Add it into the accumulator 92 adwc $0,r1 # Possible carry into high half 93 jbr ad2 # Join common code 94/* 95 * Here when the digit won't fit in the accumulator 96 */ 97ad1: incl r5 # Ignore the digit, bump exponent 98/* 99 * If we have seen a decimal point, decrease the exponent by 1 100 */ 101ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen 102 decl r5 # Decrease exponent 103ad3: 104/* 105 * Fetch the next character, back for more 106 */ 107 movzbl (r2)+,r4 # Fetch 108 jbr ad0 # Try again 109/* 110 * Not a digit. Could it be a decimal point? 111 */ 112ad4: cmpb r4,$'. # If it's not a decimal point, either it's 113 jneq ad5 # the end of the number or the start of 114 # the exponent. 115 jbcs $decpt,r3,ad3 # If it IS a decimal point, we record that 116 # we've seen one, and keep collecting 117 # digits if it is the first one. 118/* 119 * Check for an exponent 120 */ 121ad5: clrl r6 # Initialize the exponent accumulator 122 123 cmpb r4,$'e # We allow both lower case e 124 jeql ex1 # ... and ... 125 cmpb r4,$'E # upper-case E 126 jneq ex7 127/* 128 * Does the exponent have a sign? 129 */ 130ex1: movzbl (r2)+,r4 # Get next character 131 cmpb r4,$'+ # Positive sign? 132 jeql ex2 # ... yes ... 133 cmpb r4,$'- # Negative sign? 134 jneq ex3 # ... no ... 135 bisb2 $1<esign,r3 # Indicate exponent is negative 136ex2: movzbl (r2)+,r4 # Grab the next character 137/* 138 * Accumulate exponent digits in r6 139 */ 140ex3: cmpb r4,$'0 # A digit is within the range 141 jlss ex4 # '0' through 142 cmpb r4,$'9 # '9', 143 jgtr ex4 # inclusive. 144 cmpl r6,$214748364 # Exponent outrageously large already? 145 jgeq ex2 # ... yes 146 moval (r6)[r6],r6 # r6 *= 5 147 movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0' 148 jbr ex2 # Go 'round again 149ex4: 150/* 151 * Now get the final exponent and force it within a reasonable 152 * range so our scaling loops don't take forever for values 153 * that will ultimately cause overflow or underflow anyway. 154 * A tight check on over/underflow will be done by ldexp. 155 */ 156 jbc $esign,r3,ex5 # Jump if exponent not negative 157 mnegl r6,r6 # If sign, negate exponent 158ex5: addl2 r6,r5 # Add given exponent to calculated exponent 159 cmpl r5,$-100 # Absurdly small? 160 jgtr ex6 # ... no 161 movl $-100,r5 # ... yes, force within limit 162ex6: cmpl r5,$100 # Absurdly large? 163 jlss ex7 # ... no 164 movl $100,r5 # ... yes, force within bounds 165ex7: 166/* 167 * Our number has now been reduced to a mantissa and an exponent. 168 * The mantissa is a 63-bit positive binary integer in r0,r1, 169 * and the exponent is a signed power of 10 in r5. The msign 170 * bit in r3 will be on if the mantissa should ultimately be 171 * considered negative. 172 * 173 * We now have to convert it to a standard format floating point 174 * number. This will be done by accumulating a binary exponent 175 * in r2, as we progressively get r5 closer to zero. 176 * 177 * Don't bother scaling if the mantissa is zero 178 */ 179 movq r0,r0 # Mantissa zero? 180 jeql exit # ... yes 181 182 clrl r2 # Initialize binary exponent 183 tstl r5 # Which way to scale? 184 jleq sd0 # Scale down if decimal exponent <= 0 185/* 186 * Scale up by "multiplying" r0,r1 by 10 as many times as necessary, 187 * as follows: 188 * 189 * Step 1: Shift r0,r1 right as necessary to ensure that no 190 * overflow can occur when multiplying. 191 */ 192su0: cmpl r1,$429496729 # Compare high word to (2**31)/5 193 jlss su1 # Jump out if guaranteed safe 194 ashq $-1,r0,r0 # Else shift right one bit 195 incl r2 # bump exponent to compensate 196 jbr su0 # and go back to test again. 197/* 198 * Step 2: Multiply r0,r1 by 5, by appropriate shifting and 199 * double-precision addition 200 */ 201su1: ashq $2,r0,r6 # (r6,r7) := (r0,r1) * 4 202 addl2 r6,r0 # Add low-order halves 203 adwc r7,r1 # and high-order halves 204/* 205 * Step 3: Increment the binary exponent to take care of the final 206 * factor of 2, and go back if we still need to scale more. 207 */ 208 incl r2 # Increment the exponent 209 sobgtr r5,su0 # and back for more (maybe) 210 211 jbr cm0 # Merge to build final value 212 213/* 214 * Scale down. We must "divide" r0,r1 by 10 as many times 215 * as needed, as follows: 216 * 217 * Step 0: Right now, the condition codes reflect the state 218 * of r5. If it's zero, we are done. 219 */ 220sd0: jeql cm0 # If finished, build final number 221/* 222 * Step 1: Shift r0,r1 left until the high-order bit (not counting 223 * the sign bit) is nonzero, so that the division will preserve 224 * as much precision as possible. 225 */ 226 tstl r1 # Is the entire high-order half zero? 227 jneq sd2 # ...no, go shift one bit at a time 228 ashq $30,r0,r0 # ...yes, shift left 30, 229 subl2 $30,r2 # decrement the exponent to compensate, 230 # and now it's known to be safe to shift 231 # at least once more. 232sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and 233 decl r2 # decrement the exponent to compensate 234sd2: jbc $30,r1,sd1 # If the high-order bit is off, go shift 235/* 236 * Step 2: Divide the high-order part of (r0,r1) by 5, 237 * giving a quotient in r1 and a remainder in r7. 238 */ 239sd3: movl r1,r6 # Copy the high-order part 240 clrl r7 # Zero-extend to 64 bits 241 ediv $5,r6,r1,r7 # Divide (cannot overflow) 242/* 243 * Step 3: Divide the low-order part of (r0,r1) by 5, 244 * using the remainder from step 2 for rounding. 245 * Note that the result of this computation is unsigned, 246 * so we have to allow for the fact that an ordinary division 247 * by 5 could overflow. We make allowance by dividing by 10, 248 * multiplying the quotient by 2, and using the remainder 249 * to adjust the modified quotient. 250 */ 251 addl3 $2,r0,r6 # Dividend is low part of (r0,r1) plus 252 adwc $0,r7 # 2 for rounding plus 253 # (2**32) * previous remainder 254 ediv $10,r6,r0,r6 # r0 := quotient, r6 := remainder. 255 addl2 r0,r0 # Make r0 result of dividing by 5 256 cmpl r6,$5 # If remainder is 5 or greater, 257 jlss sd4 # increment the adjustted quotient. 258 incl r0 259/* 260 * Step 4: Increment the decimal exponent, decrement the binary 261 * exponent (to make the division by 5 into a division by 10), 262 * and back for another iteration. 263 */ 264sd4: decl r2 # Binary exponent 265 aoblss $0,r5,sd2 266/* 267 * We now have the following: 268 * 269 * r0: low-order half of a 64-bit integer 270 * r1: high-order half of the same 64-bit integer 271 * r2: a binary exponent 272 * 273 * Our final result is the integer represented by (r0,r1) 274 * multiplied by 2 to the power contained in r2. 275 * We will transform (r0,r1) into a floating-point value, 276 * set the sign appropriately, and let ldexp do the 277 * rest of the work. 278 * 279 * Step 1: if the high-order bit (excluding the sign) of 280 * the high-order half (r1) is 1, then we have 63 bits of 281 * fraction, too many to convert easily. However, we also 282 * know we won't need them all, so we will just throw the 283 * low-order bit away (and adjust the exponent appropriately). 284 */ 285cm0: jbc $30,r1,cm1 # jump if no adjustment needed 286 ashq $-1,r0,r0 # lose the low-order bit 287 incl r2 # increase the exponent to compensate 288/* 289 * Step 2: split the 62-bit number in (r0,r1) into two 290 * 31-bit positive quantities 291 */ 292cm1: ashq $1,r0,r0 # put the high-order bits in r1 293 # and a 0 in the bottom of r0 294 rotl $-1,r0,r0 # right-justify the bits in r0 295 # moving the 0 from the ashq 296 # into the sign bit. 297/* 298 * Step 3: convert both halves to floating point 299 */ 300 cvtld r0,r6 # low-order part in r6-r7 301 cvtld r1,r0 # high-order part in r0-r1 302/* 303 * Step 4: multiply the high order part by 2**31 and combine them 304 */ 305 muld2 two31,r0 # multiply 306 addd2 r6,r0 # combine 307/* 308 * Step 5: if appropriate, negate the floating value 309 */ 310 jbc $msign,r3,cm2 # Jump if mantissa not signed 311 mnegd r0,r0 # If negative, make it so 312/* 313 * Step 6: call ldexp to complete the job 314 */ 315cm2: pushl r2 # Put exponent in parameter list 316 movd r0,-(sp) # and also mantissa 317 calls $3,_ldexp # go combine them 318 319exit: 320 ret 321 322 .align 2 323two31: .word 0x5000 # 2 ** 31 324 .word 0 # (=2147483648) 325 .word 0 # in floating-point 326 .word 0 # (so atof doesn't have to convert it) 327