1/* 2 * Copyright (c) 1980 Regents of the University of California. 3 * All rights reserved. The Berkeley software License Agreement 4 * specifies the terms and conditions for redistribution. 5 */ 6 7#ifndef lint 8 .asciz "@(#)atof.s 5.2 (Berkeley) 06/05/85" 9#endif not lint 10 11#include "DEFS.h" 12 13/* 14 * atof: convert ascii to floating 15 * 16 * C usage: 17 * 18 * double atof (s) 19 * char *s; 20 * 21 * Register usage: 22 * 23 * r0-1: value being developed 24 * r2: first section: pointer to the next character 25 * second section: binary exponent 26 * r3: flags 27 * r4: first section: the current character 28 * second section: scratch 29 * r5: the decimal exponent 30 * r6-7: scratch 31 */ 32 .set msign,0 # mantissa has negative sign 33 .set esign,1 # exponent has negative sign 34 .set decpt,2 # decimal point encountered 35 36ENTRY(atof, R6|R7) 37/* 38 * Initialization 39 */ 40 clrl r3 # All flags start out false 41 movl 4(ap),r2 # Address the first character 42 clrl r5 # Clear starting exponent 43/* 44 * Skip leading white space 45 */ 46sk0: movzbl (r2)+,r4 # Fetch the next (first) character 47 cmpb $' ,r4 # Is it blank? 48 jeql sk0 # ...yes 49 cmpb r4,$8 # 8 is lowest of white-space group 50 jlss sk1 # Jump if char too low to be white space 51 cmpb r4,$13 # 13 is highest of white-space group 52 jleq sk0 # Jump if character is white space 53sk1: 54/* 55 * Check for a sign 56 */ 57 cmpb $'+,r4 # Positive sign? 58 jeql cs1 # ... yes 59 cmpb $'-,r4 # Negative sign? 60 jneq cs2 # ... no 61 bisb2 $1<msign,r3 # Indicate a negative mantissa 62cs1: movzbl (r2)+,r4 # Skip the character 63cs2: 64/* 65 * Accumulate digits, keeping track of the exponent 66 */ 67 clrq r0 # Clear the accumulator 68ad0: cmpb r4,$'0 # Do we have a digit? 69 jlss ad4 # ... no, too small 70 cmpb r4,$'9 71 jgtr ad4 # ... no, too large 72/* 73 * We got a digit. Accumulate it 74 */ 75 cmpl r1,$214748364 # Would this digit cause overflow? 76 jgeq ad1 # ... yes 77/* 78 * Multiply (r0,r1) by 10. This is done by developing 79 * (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits, 80 * and adding the two quadwords. 81 */ 82 ashq $1,r0,r6 # (r6,r7)=(r0,r1)*2 83 ashq $3,r0,r0 # (r0,r1)=(r0,r1)*8 84 addl2 r6,r0 # Add low halves 85 adwc r7,r1 # Add high halves 86/* 87 * Add in the digit 88 */ 89 subl2 $'0,r4 # Get the digit value 90 addl2 r4,r0 # Add it into the accumulator 91 adwc $0,r1 # Possible carry into high half 92 jbr ad2 # Join common code 93/* 94 * Here when the digit won't fit in the accumulator 95 */ 96ad1: incl r5 # Ignore the digit, bump exponent 97/* 98 * If we have seen a decimal point, decrease the exponent by 1 99 */ 100ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen 101 decl r5 # Decrease exponent 102ad3: 103/* 104 * Fetch the next character, back for more 105 */ 106 movzbl (r2)+,r4 # Fetch 107 jbr ad0 # Try again 108/* 109 * Not a digit. Could it be a decimal point? 110 */ 111ad4: cmpb r4,$'. # If it's not a decimal point, either it's 112 jneq ad5 # the end of the number or the start of 113 # the exponent. 114 jbcs $decpt,r3,ad3 # If it IS a decimal point, we record that 115 # we've seen one, and keep collecting 116 # digits if it is the first one. 117/* 118 * Check for an exponent 119 */ 120ad5: clrl r6 # Initialize the exponent accumulator 121 122 cmpb r4,$'e # We allow both lower case e 123 jeql ex1 # ... and ... 124 cmpb r4,$'E # upper-case E 125 jneq ex7 126/* 127 * Does the exponent have a sign? 128 */ 129ex1: movzbl (r2)+,r4 # Get next character 130 cmpb r4,$'+ # Positive sign? 131 jeql ex2 # ... yes ... 132 cmpb r4,$'- # Negative sign? 133 jneq ex3 # ... no ... 134 bisb2 $1<esign,r3 # Indicate exponent is negative 135ex2: movzbl (r2)+,r4 # Grab the next character 136/* 137 * Accumulate exponent digits in r6 138 */ 139ex3: cmpb r4,$'0 # A digit is within the range 140 jlss ex4 # '0' through 141 cmpb r4,$'9 # '9', 142 jgtr ex4 # inclusive. 143 cmpl r6,$214748364 # Exponent outrageously large already? 144 jgeq ex2 # ... yes 145 moval (r6)[r6],r6 # r6 *= 5 146 movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0' 147 jbr ex2 # Go 'round again 148ex4: 149/* 150 * Now get the final exponent and force it within a reasonable 151 * range so our scaling loops don't take forever for values 152 * that will ultimately cause overflow or underflow anyway. 153 * A tight check on over/underflow will be done by ldexp. 154 */ 155 jbc $esign,r3,ex5 # Jump if exponent not negative 156 mnegl r6,r6 # If sign, negate exponent 157ex5: addl2 r6,r5 # Add given exponent to calculated exponent 158 cmpl r5,$-100 # Absurdly small? 159 jgtr ex6 # ... no 160 movl $-100,r5 # ... yes, force within limit 161ex6: cmpl r5,$100 # Absurdly large? 162 jlss ex7 # ... no 163 movl $100,r5 # ... yes, force within bounds 164ex7: 165/* 166 * Our number has now been reduced to a mantissa and an exponent. 167 * The mantissa is a 63-bit positive binary integer in r0,r1, 168 * and the exponent is a signed power of 10 in r5. The msign 169 * bit in r3 will be on if the mantissa should ultimately be 170 * considered negative. 171 * 172 * We now have to convert it to a standard format floating point 173 * number. This will be done by accumulating a binary exponent 174 * in r2, as we progressively get r5 closer to zero. 175 * 176 * Don't bother scaling if the mantissa is zero 177 */ 178 movq r0,r0 # Mantissa zero? 179 jeql exit # ... yes 180 181 clrl r2 # Initialize binary exponent 182 tstl r5 # Which way to scale? 183 jleq sd0 # Scale down if decimal exponent <= 0 184/* 185 * Scale up by "multiplying" r0,r1 by 10 as many times as necessary, 186 * as follows: 187 * 188 * Step 1: Shift r0,r1 right as necessary to ensure that no 189 * overflow can occur when multiplying. 190 */ 191su0: cmpl r1,$429496729 # Compare high word to (2**31)/5 192 jlss su1 # Jump out if guaranteed safe 193 ashq $-1,r0,r0 # Else shift right one bit 194 incl r2 # bump exponent to compensate 195 jbr su0 # and go back to test again. 196/* 197 * Step 2: Multiply r0,r1 by 5, by appropriate shifting and 198 * double-precision addition 199 */ 200su1: ashq $2,r0,r6 # (r6,r7) := (r0,r1) * 4 201 addl2 r6,r0 # Add low-order halves 202 adwc r7,r1 # and high-order halves 203/* 204 * Step 3: Increment the binary exponent to take care of the final 205 * factor of 2, and go back if we still need to scale more. 206 */ 207 incl r2 # Increment the exponent 208 sobgtr r5,su0 # and back for more (maybe) 209 210 jbr cm0 # Merge to build final value 211 212/* 213 * Scale down. We must "divide" r0,r1 by 10 as many times 214 * as needed, as follows: 215 * 216 * Step 0: Right now, the condition codes reflect the state 217 * of r5. If it's zero, we are done. 218 */ 219sd0: jeql cm0 # If finished, build final number 220/* 221 * Step 1: Shift r0,r1 left until the high-order bit (not counting 222 * the sign bit) is nonzero, so that the division will preserve 223 * as much precision as possible. 224 */ 225 tstl r1 # Is the entire high-order half zero? 226 jneq sd2 # ...no, go shift one bit at a time 227 ashq $30,r0,r0 # ...yes, shift left 30, 228 subl2 $30,r2 # decrement the exponent to compensate, 229 # and now it's known to be safe to shift 230 # at least once more. 231sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and 232 decl r2 # decrement the exponent to compensate 233sd2: jbc $30,r1,sd1 # If the high-order bit is off, go shift 234/* 235 * Step 2: Divide the high-order part of (r0,r1) by 5, 236 * giving a quotient in r1 and a remainder in r7. 237 */ 238sd3: movl r1,r6 # Copy the high-order part 239 clrl r7 # Zero-extend to 64 bits 240 ediv $5,r6,r1,r7 # Divide (cannot overflow) 241/* 242 * Step 3: Divide the low-order part of (r0,r1) by 5, 243 * using the remainder from step 2 for rounding. 244 * Note that the result of this computation is unsigned, 245 * so we have to allow for the fact that an ordinary division 246 * by 5 could overflow. We make allowance by dividing by 10, 247 * multiplying the quotient by 2, and using the remainder 248 * to adjust the modified quotient. 249 */ 250 addl3 $2,r0,r6 # Dividend is low part of (r0,r1) plus 251 adwc $0,r7 # 2 for rounding plus 252 # (2**32) * previous remainder 253 ediv $10,r6,r0,r6 # r0 := quotient, r6 := remainder. 254 addl2 r0,r0 # Make r0 result of dividing by 5 255 cmpl r6,$5 # If remainder is 5 or greater, 256 jlss sd4 # increment the adjustted quotient. 257 incl r0 258/* 259 * Step 4: Increment the decimal exponent, decrement the binary 260 * exponent (to make the division by 5 into a division by 10), 261 * and back for another iteration. 262 */ 263sd4: decl r2 # Binary exponent 264 aoblss $0,r5,sd2 265/* 266 * We now have the following: 267 * 268 * r0: low-order half of a 64-bit integer 269 * r1: high-order half of the same 64-bit integer 270 * r2: a binary exponent 271 * 272 * Our final result is the integer represented by (r0,r1) 273 * multiplied by 2 to the power contained in r2. 274 * We will transform (r0,r1) into a floating-point value, 275 * set the sign appropriately, and let ldexp do the 276 * rest of the work. 277 * 278 * Step 1: if the high-order bit (excluding the sign) of 279 * the high-order half (r1) is 1, then we have 63 bits of 280 * fraction, too many to convert easily. However, we also 281 * know we won't need them all, so we will just throw the 282 * low-order bit away (and adjust the exponent appropriately). 283 */ 284cm0: jbc $30,r1,cm1 # jump if no adjustment needed 285 ashq $-1,r0,r0 # lose the low-order bit 286 incl r2 # increase the exponent to compensate 287/* 288 * Step 2: split the 62-bit number in (r0,r1) into two 289 * 31-bit positive quantities 290 */ 291cm1: ashq $1,r0,r0 # put the high-order bits in r1 292 # and a 0 in the bottom of r0 293 rotl $-1,r0,r0 # right-justify the bits in r0 294 # moving the 0 from the ashq 295 # into the sign bit. 296/* 297 * Step 3: convert both halves to floating point 298 */ 299 cvtld r0,r6 # low-order part in r6-r7 300 cvtld r1,r0 # high-order part in r0-r1 301/* 302 * Step 4: multiply the high order part by 2**31 and combine them 303 */ 304 muld2 two31,r0 # multiply 305 addd2 r6,r0 # combine 306/* 307 * Step 5: if appropriate, negate the floating value 308 */ 309 jbc $msign,r3,cm2 # Jump if mantissa not signed 310 mnegd r0,r0 # If negative, make it so 311/* 312 * Step 6: call ldexp to complete the job 313 */ 314cm2: pushl r2 # Put exponent in parameter list 315 movd r0,-(sp) # and also mantissa 316 calls $3,_ldexp # go combine them 317 318exit: 319 ret 320 321 .align 2 322two31: .word 0x5000 # 2 ** 31 323 .word 0 # (=2147483648) 324 .word 0 # in floating-point 325 .word 0 # (so atof doesn't have to convert it) 326