1/* 2 * Copyright (c) 1988 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 5.2 (Berkeley) 08/02/88" 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(fp),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 incl r2 59 cmpb $' ,r4 # Is it blank? 60 beql sk0 # ...yes 61 cmpb r4,$8 # 8 is lowest of white-space group 62 blss sk1 # Jump if char too low to be white space 63 cmpb r4,$13 # 13 is highest of white-space group 64 bleq sk0 # Jump if character is white space 65sk1: 66/* 67 * Check for a sign 68 */ 69 cmpb $'+,r4 # Positive sign? 70 beql cs1 # ... yes 71 cmpb $'-,r4 # Negative sign? 72 bneq cs2 # ... no 73 orb2 $1<msign,r3 # Indicate a negative mantissa 74cs1: movzbl (r2),r4 # Skip the character 75 incl r2 76cs2: 77/* 78 * Accumulate digits, keeping track of the exponent 79 */ 80 clrl r1 81 clrl r0 # Clear the accumulator 82ad0: cmpb r4,$'0 # Do we have a digit? 83 blss ad4 # ... no, too small 84 cmpb r4,$'9 85 bgtr ad4 # ... no, too large 86/* 87 * We got a digit. Accumulate it 88 */ 89 cmpl r0,$214748364 # Would this digit cause overflow? 90 bgeq ad1 # ... yes 91/* 92 * Multiply (r0,r1) by 10. This is done by developing 93 * (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits, 94 * and adding the two quadwords. 95 */ 96 shlq $1,r0,r6 # (r6,r7)=(r0,r1)*2 97 shlq $3,r0,r0 # (r0,r1)=(r0,r1)*8 98 addl2 r7,r1 # Add low halves 99 adwc r6,r0 # Add high halves 100/* 101 * Add in the digit 102 */ 103 subl2 $'0,r4 # Get the digit value 104 addl2 r4,r1 # Add it into the accumulator 105 adwc $0,r0 # Possible carry into high half 106 brb ad2 # Join common code 107/* 108 * Here when the digit won't fit in the accumulator 109 */ 110ad1: incl r5 # Ignore the digit, bump exponent 111/* 112 * If we have seen a decimal point, decrease the exponent by 1 113 */ 114ad2: bbc $decpt,r3,ad3 # Jump if decimal point not seen 115 decl r5 # Decrease exponent 116ad3: 117/* 118 * Fetch the next character, back for more 119 */ 120 movzbl (r2),r4 # Fetch 121 incl r2 122 brb ad0 # Try again 123/* 124 * Not a digit. Could it be a decimal point? 125 */ 126ad4: cmpb r4,$'. # If it's not a decimal point, either it's 127 bneq ad5 # the end of the number or the start of 128 # the exponent. 129 bbs $decpt,r3,ad5 130 orb2 $1<decpt,r3 # If it IS a decimal point, we record that 131 brb ad3 # we've seen one, and keep collecting 132 # digits if it is the first one. 133 134/* 135 * Check for an exponent 136 */ 137ad5: clrl r6 # Initialize the exponent accumulator 138 139 cmpb r4,$'e # We allow both lower case e 140 beql ex1 # ... and ... 141 cmpb r4,$'E # upper-case E 142 bneq ex7 143/* 144 * Does the exponent have a sign? 145 */ 146ex1: movzbl (r2),r4 # Get next character 147 incl r2 148 cmpb r4,$'+ # Positive sign? 149 beql ex2 # ... yes ... 150 cmpb r4,$'- # Negative sign? 151 bneq ex3 # ... no ... 152 orb2 $1<esign,r3 # Indicate exponent is negative 153ex2: movzbl (r2),r4 # Grab the next character 154 incl r2 155/* 156 * Accumulate exponent digits in r6 157 */ 158ex3: cmpb r4,$'0 # A digit is within the range 159 blss ex4 # '0' through 160 cmpb r4,$'9 # '9', 161 bgtr ex4 # inclusive. 162 cmpl r6,$214748364 # Exponent outrageously large already? 163 bgeq ex2 # ... yes 164 moval (r6)[r6],r6 # r6 *= 5 165 movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0' 166 brb ex2 # Go 'round again 167ex4: 168/* 169 * Now get the final exponent and force it within a reasonable 170 * range so our scaling loops don't take forever for values 171 * that will ultimately cause overflow or underflow anyway. 172 * A tight check on over/underflow will be done by ldexp. 173 */ 174 bbc $esign,r3,ex5 # Jump if exponent not negative 175 mnegl r6,r6 # If sign, negate exponent 176ex5: addl2 r6,r5 # Add given exponent to calculated exponent 177 cmpl r5,$-100 # Absurdly small? 178 bgtr ex6 # ... no 179 movl $-100,r5 # ... yes, force within limit 180ex6: cmpl r5,$100 # Absurdly large? 181 blss ex7 # ... no 182 movl $100,r5 # ... yes, force within bounds 183ex7: 184/* 185 * Our number has now been reduced to a mantissa and an exponent. 186 * The mantissa is a 63-bit positive binary integer in r0,r1, 187 * and the exponent is a signed power of 10 in r5. The msign 188 * bit in r3 will be on if the mantissa should ultimately be 189 * considered negative. 190 * 191 * We now have to convert it to a standard format floating point 192 * number. This will be done by accumulating a binary exponent 193 * in r2, as we progressively get r5 closer to zero. 194 * 195 * Don't bother scaling if the mantissa is zero 196 */ 197 tstl r1 198 bneq 1f 199 tstl r0 # Mantissa zero? 200 jeql exit # ... yes 201 2021: clrl r2 # Initialize binary exponent 203 tstl r5 # Which way to scale? 204 bleq sd0 # Scale down if decimal exponent <= 0 205/* 206 * Scale up by "multiplying" r0,r1 by 10 as many times as necessary, 207 * as follows: 208 * 209 * Step 1: Shift r0,r1 right as necessary to ensure that no 210 * overflow can occur when multiplying. 211 */ 212su0: cmpl r0,$429496729 # Compare high word to (2**31)/5 213 blss su1 # Jump out if guaranteed safe 214 shrq $1,r0,r0 # Else shift right one bit 215 incl r2 # bump exponent to compensate 216 brb su0 # and go back to test again. 217/* 218 * Step 2: Multiply r0,r1 by 5, by appropriate shifting and 219 * double-precision addition 220 */ 221su1: shlq $2,r0,r6 # (r6,r7) := (r0,r1) * 4 222 addl2 r7,r1 # Add low-order halves 223 adwc r6,r0 # and high-order halves 224/* 225 * Step 3: Increment the binary exponent to take care of the final 226 * factor of 2, and go back if we still need to scale more. 227 */ 228 incl r2 # Increment the exponent 229 decl r5 # ...sobgtr r5,su0 230 bgtr su0 # and back for more (maybe) 231 232 brb cm0 # Merge to build final value 233 234/* 235 * Scale down. We must "divide" r0,r1 by 10 as many times 236 * as needed, as follows: 237 * 238 * Step 0: Right now, the condition codes reflect the state 239 * of r5. If it's zero, we are done. 240 */ 241sd0: beql cm0 # If finished, build final number 242/* 243 * Step 1: Shift r0,r1 left until the high-order bit (not counting 244 * the sign bit) is nonzero, so that the division will preserve 245 * as much precision as possible. 246 */ 247 tstl r0 # Is the entire high-order half zero? 248 bneq sd2 # ...no, go shift one bit at a time 249 shlq $30,r0,r0 # ...yes, shift left 30, 250 subl2 $30,r2 # decrement the exponent to compensate, 251 # and now it's known to be safe to shift 252 # at least once more. 253sd1: shlq $1,r0,r0 # Shift (r0,r1) left one, and 254 decl r2 # decrement the exponent to compensate 255sd2: bbc $30,r0,sd1 # If the high-order bit is off, go shift 256/* 257 * Step 2: Divide the high-order part of (r0,r1) by 5, 258 * giving a quotient in r1 and a remainder in r7. 259 */ 260sd3: movl r0,r7 # Copy the high-order part 261 clrl r6 # Zero-extend to 64 bits 262 ediv $5,r6,r0,r6 # Divide (cannot overflow) 263/* 264 * Step 3: Divide the low-order part of (r0,r1) by 5, 265 * using the remainder from step 2 for rounding. 266 * Note that the result of this computation is unsigned, 267 * so we have to allow for the fact that an ordinary division 268 * by 5 could overflow. We make allowance by dividing by 10, 269 * multiplying the quotient by 2, and using the remainder 270 * to adjust the modified quotient. 271 */ 272 addl3 $2,r1,r7 # Dividend is low part of (r0,r1) plus 273 adwc $0,r6 # 2 for rounding plus 274 # (2**32) * previous remainder 275 ediv $10,r6,r1,r7 # r1 := quotient, r7 := remainder. 276 addl2 r1,r1 # Make r1 result of dividing by 5 277 cmpl r7,$5 # If remainder is 5 or greater, 278 blss sd4 # increment the adjustted quotient. 279 incl r1 280/* 281 * Step 4: Increment the decimal exponent, decrement the binary 282 * exponent (to make the division by 5 into a division by 10), 283 * and back for another iteration. 284 */ 285sd4: decl r2 # Binary exponent 286 aoblss $0,r5,sd2 287/* 288 * We now have the following: 289 * 290 * r0: high-order half of a 64-bit integer 291 * r1: load-order half of the same 64-bit integer 292 * r2: a binary exponent 293 * 294 * Our final result is the integer represented by (r0,r1) 295 * multiplied by 2 to the power contained in r2. 296 * We will transform (r0,r1) into a floating-point value, 297 * set the sign appropriately, and let ldexp do the 298 * rest of the work. 299 * 300 * Step 1: if the high-order bit (excluding the sign) of 301 * the high-order half (r0) is 1, then we have 63 bits of 302 * fraction, too many to convert easily. However, we also 303 * know we won't need them all, so we will just throw the 304 * low-order bit away (and adjust the exponent appropriately). 305 */ 306cm0: bbc $30,r0,cm1 # jump if no adjustment needed 307 shrq $1,r0,r0 # lose the low-order bit 308 incl r2 # increase the exponent to compensate 309/* 310 * Step 2: split the 62-bit number in (r0,r1) into two 311 * 31-bit positive quantities 312 */ 313cm1: shlq $1,r0,r0 # put the high-order bits in r0 314 # and a 0 in the bottom of r1 315 shrl $1,r1,r1 # right-justify the bits in r1 316 # moving 0 into the sign bit. 317/* 318 * Step 3: convert both halves to floating point 319 */ 320 cvld r1 321 std r6 # low-order part in r6-r7 322 cvld r0 323 std r0 # high-order part in r0-r1 324/* 325 * Step 4: multiply the high order part by 2**31 and combine them 326 */ 327 ldd two31 328 muld r0 # multiply 329 addd r6 # combine 330/* 331 * Step 5: if appropriate, negate the floating value 332 */ 333 bbc $msign,r3,cm2 # Jump if mantissa not signed 334 negd # If negative, make it so 335/* 336 * Step 6: call ldexp to complete the job 337 */ 338cm2: pushl r2 # Put exponent in parameter list 339 pushd # and also mantissa 340 calls $3,_ldexp # go combine them 341 342exit: 343 ret 344 345 .align 2 346two31: .long 0x50000000 # (=2147483648) 2 ** 31 in floating-point 347 .long 0 # so atof doesn't have to convert it 348