1/* atof.s 4.3 84/11/01 */ 2 3#include "DEFS.h" 4 5/* 6 * atof: convert ascii to floating 7 * 8 * C usage: 9 * 10 * double atof (s) 11 * char *s; 12 * 13 * Register usage: 14 * 15 * r0-1: value being developed 16 * r2: first section: pointer to the next character 17 * second section: binary exponent 18 * r3: flags 19 * r4: first section: the current character 20 * second section: scratch 21 * r5: the decimal exponent 22 * r6-7: scratch 23 */ 24 .set msign,0 # mantissa has negative sign 25 .set esign,1 # exponent has negative sign 26 .set decpt,2 # decimal point encountered 27 28ENTRY(atof, R6|R7) 29/* 30 * Initialization 31 */ 32 clrl r3 # All flags start out false 33 movl 4(ap),r2 # Address the first character 34 clrl r5 # Clear starting exponent 35/* 36 * Skip leading white space 37 */ 38sk0: movzbl (r2)+,r4 # Fetch the next (first) character 39 cmpb $' ,r4 # Is it blank? 40 jeql sk0 # ...yes 41 cmpb r4,$8 # 8 is lowest of white-space group 42 jlss sk1 # Jump if char too low to be white space 43 cmpb r4,$13 # 13 is highest of white-space group 44 jleq sk0 # Jump if character is white space 45sk1: 46/* 47 * Check for a sign 48 */ 49 cmpb $'+,r4 # Positive sign? 50 jeql cs1 # ... yes 51 cmpb $'-,r4 # Negative sign? 52 jneq cs2 # ... no 53 bisb2 $1<msign,r3 # Indicate a negative mantissa 54cs1: movzbl (r2)+,r4 # Skip the character 55cs2: 56/* 57 * Accumulate digits, keeping track of the exponent 58 */ 59 clrq r0 # Clear the accumulator 60ad0: cmpb r4,$'0 # Do we have a digit? 61 jlss ad4 # ... no, too small 62 cmpb r4,$'9 63 jgtr ad4 # ... no, too large 64/* 65 * We got a digit. Accumulate it 66 */ 67 cmpl r1,$214748364 # Would this digit cause overflow? 68 jgeq ad1 # ... yes 69/* 70 * Multiply (r0,r1) by 10. This is done by developing 71 * (r0,r1)*2 in (r6,r7), shifting (r0,r1) left three bits, 72 * and adding the two quadwords. 73 */ 74 ashq $1,r0,r6 # (r6,r7)=(r0,r1)*2 75 ashq $3,r0,r0 # (r0,r1)=(r0,r1)*8 76 addl2 r6,r0 # Add low halves 77 adwc r7,r1 # Add high halves 78/* 79 * Add in the digit 80 */ 81 subl2 $'0,r4 # Get the digit value 82 addl2 r4,r0 # Add it into the accumulator 83 adwc $0,r1 # Possible carry into high half 84 jbr ad2 # Join common code 85/* 86 * Here when the digit won't fit in the accumulator 87 */ 88ad1: incl r5 # Ignore the digit, bump exponent 89/* 90 * If we have seen a decimal point, decrease the exponent by 1 91 */ 92ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen 93 decl r5 # Decrease exponent 94ad3: 95/* 96 * Fetch the next character, back for more 97 */ 98 movzbl (r2)+,r4 # Fetch 99 jbr ad0 # Try again 100/* 101 * Not a digit. Could it be a decimal point? 102 */ 103ad4: cmpb r4,$'. # If it's not a decimal point, either it's 104 jneq ad5 # the end of the number or the start of 105 # the exponent. 106 jbcs $decpt,r3,ad3 # If it IS a decimal point, we record that 107 # we've seen one, and keep collecting 108 # digits if it is the first one. 109/* 110 * Check for an exponent 111 */ 112ad5: clrl r6 # Initialize the exponent accumulator 113 114 cmpb r4,$'e # We allow both lower case e 115 jeql ex1 # ... and ... 116 cmpb r4,$'E # upper-case E 117 jneq ex7 118/* 119 * Does the exponent have a sign? 120 */ 121ex1: movzbl (r2)+,r4 # Get next character 122 cmpb r4,$'+ # Positive sign? 123 jeql ex2 # ... yes ... 124 cmpb r4,$'- # Negative sign? 125 jneq ex3 # ... no ... 126 bisb2 $1<esign,r3 # Indicate exponent is negative 127ex2: movzbl (r2)+,r4 # Grab the next character 128/* 129 * Accumulate exponent digits in r6 130 */ 131ex3: cmpb r4,$'0 # A digit is within the range 132 jlss ex4 # '0' through 133 cmpb r4,$'9 # '9', 134 jgtr ex4 # inclusive. 135 cmpl r6,$214748364 # Exponent outrageously large already? 136 jgeq ex2 # ... yes 137 moval (r6)[r6],r6 # r6 *= 5 138 movaw -'0(r4)[r6],r6 # r6 = r6 * 2 + r4 - '0' 139 jbr ex2 # Go 'round again 140ex4: 141/* 142 * Now get the final exponent and force it within a reasonable 143 * range so our scaling loops don't take forever for values 144 * that will ultimately cause overflow or underflow anyway. 145 * A tight check on over/underflow will be done by ldexp. 146 */ 147 jbc $esign,r3,ex5 # Jump if exponent not negative 148 mnegl r6,r6 # If sign, negate exponent 149ex5: addl2 r6,r5 # Add given exponent to calculated exponent 150 cmpl r5,$-100 # Absurdly small? 151 jgtr ex6 # ... no 152 movl $-100,r5 # ... yes, force within limit 153ex6: cmpl r5,$100 # Absurdly large? 154 jlss ex7 # ... no 155 movl $100,r5 # ... yes, force within bounds 156ex7: 157/* 158 * Our number has now been reduced to a mantissa and an exponent. 159 * The mantissa is a 63-bit positive binary integer in r0,r1, 160 * and the exponent is a signed power of 10 in r5. The msign 161 * bit in r3 will be on if the mantissa should ultimately be 162 * considered negative. 163 * 164 * We now have to convert it to a standard format floating point 165 * number. This will be done by accumulating a binary exponent 166 * in r2, as we progressively get r5 closer to zero. 167 * 168 * Don't bother scaling if the mantissa is zero 169 */ 170 movq r0,r0 # Mantissa zero? 171 jeql exit # ... yes 172 173 clrl r2 # Initialize binary exponent 174 tstl r5 # Which way to scale? 175 jleq sd0 # Scale down if decimal exponent <= 0 176/* 177 * Scale up by "multiplying" r0,r1 by 10 as many times as necessary, 178 * as follows: 179 * 180 * Step 1: Shift r0,r1 right as necessary to ensure that no 181 * overflow can occur when multiplying. 182 */ 183su0: cmpl r1,$429496729 # Compare high word to (2**31)/5 184 jlss su1 # Jump out if guaranteed safe 185 ashq $-1,r0,r0 # Else shift right one bit 186 incl r2 # bump exponent to compensate 187 jbr su0 # and go back to test again. 188/* 189 * Step 2: Multiply r0,r1 by 5, by appropriate shifting and 190 * double-precision addition 191 */ 192su1: ashq $2,r0,r6 # (r6,r7) := (r0,r1) * 4 193 addl2 r6,r0 # Add low-order halves 194 adwc r7,r1 # and high-order halves 195/* 196 * Step 3: Increment the binary exponent to take care of the final 197 * factor of 2, and go back if we still need to scale more. 198 */ 199 incl r2 # Increment the exponent 200 sobgtr r5,su0 # and back for more (maybe) 201 202 jbr cm0 # Merge to build final value 203 204/* 205 * Scale down. We must "divide" r0,r1 by 10 as many times 206 * as needed, as follows: 207 * 208 * Step 0: Right now, the condition codes reflect the state 209 * of r5. If it's zero, we are done. 210 */ 211sd0: jeql cm0 # If finished, build final number 212/* 213 * Step 1: Shift r0,r1 left until the high-order bit (not counting 214 * the sign bit) is nonzero, so that the division will preserve 215 * as much precision as possible. 216 */ 217 tstl r1 # Is the entire high-order half zero? 218 jneq sd2 # ...no, go shift one bit at a time 219 ashq $30,r0,r0 # ...yes, shift left 30, 220 subl2 $30,r2 # decrement the exponent to compensate, 221 # and now it's known to be safe to shift 222 # at least once more. 223sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and 224 decl r2 # decrement the exponent to compensate 225sd2: jbc $30,r1,sd1 # If the high-order bit is off, go shift 226/* 227 * Step 2: Divide the high-order part of (r0,r1) by 5, 228 * giving a quotient in r1 and a remainder in r7. 229 */ 230sd3: movl r1,r6 # Copy the high-order part 231 clrl r7 # Zero-extend to 64 bits 232 ediv $5,r6,r1,r7 # Divide (cannot overflow) 233/* 234 * Step 3: Divide the low-order part of (r0,r1) by 5, 235 * using the remainder from step 2 for rounding. 236 * Note that the result of this computation is unsigned, 237 * so we have to allow for the fact that an ordinary division 238 * by 5 could overflow. We make allowance by dividing by 10, 239 * multiplying the quotient by 2, and using the remainder 240 * to adjust the modified quotient. 241 */ 242 addl3 $2,r0,r6 # Dividend is low part of (r0,r1) plus 243 adwc $0,r7 # 2 for rounding plus 244 # (2**32) * previous remainder 245 ediv $10,r6,r0,r6 # r0 := quotient, r6 := remainder. 246 addl2 r0,r0 # Make r0 result of dividing by 5 247 cmpl r6,$5 # If remainder is 5 or greater, 248 jlss sd4 # increment the adjustted quotient. 249 incl r0 250/* 251 * Step 4: Increment the decimal exponent, decrement the binary 252 * exponent (to make the division by 5 into a division by 10), 253 * and back for another iteration. 254 */ 255sd4: decl r2 # Binary exponent 256 aoblss $0,r5,sd2 257/* 258 * We now have the following: 259 * 260 * r0: low-order half of a 64-bit integer 261 * r1: high-order half of the same 64-bit integer 262 * r2: a binary exponent 263 * 264 * Our final result is the integer represented by (r0,r1) 265 * multiplied by 2 to the power contained in r2. 266 * We will transform (r0,r1) into a floating-point value, 267 * set the sign appropriately, and let ldexp do the 268 * rest of the work. 269 * 270 * Step 1: if the high-order bit (excluding the sign) of 271 * the high-order half (r1) is 1, then we have 63 bits of 272 * fraction, too many to convert easily. However, we also 273 * know we won't need them all, so we will just throw the 274 * low-order bit away (and adjust the exponent appropriately). 275 */ 276cm0: jbc $30,r1,cm1 # jump if no adjustment needed 277 ashq $-1,r0,r0 # lose the low-order bit 278 incl r2 # increase the exponent to compensate 279/* 280 * Step 2: split the 62-bit number in (r0,r1) into two 281 * 31-bit positive quantities 282 */ 283cm1: ashq $1,r0,r0 # put the high-order bits in r1 284 # and a 0 in the bottom of r0 285 rotl $-1,r0,r0 # right-justify the bits in r0 286 # moving the 0 from the ashq 287 # into the sign bit. 288/* 289 * Step 3: convert both halves to floating point 290 */ 291 cvtld r0,r6 # low-order part in r6-r7 292 cvtld r1,r0 # high-order part in r0-r1 293/* 294 * Step 4: multiply the high order part by 2**31 and combine them 295 */ 296 muld2 two31,r0 # multiply 297 addd2 r6,r0 # combine 298/* 299 * Step 5: if appropriate, negate the floating value 300 */ 301 jbc $msign,r3,cm2 # Jump if mantissa not signed 302 mnegd r0,r0 # If negative, make it so 303/* 304 * Step 6: call ldexp to complete the job 305 */ 306cm2: pushl r2 # Put exponent in parameter list 307 movd r0,-(sp) # and also mantissa 308 calls $3,_ldexp # go combine them 309 310exit: 311 ret 312 313 .align 2 314two31: .word 0x5000 # 2 ** 31 315 .word 0 # (=2147483648) 316 .word 0 # in floating-point 317 .word 0 # (so atof doesn't have to convert it) 318