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1.1 ! root 1: # ! 2: # atof: convert ascii to floating ! 3: # ! 4: # C usage: ! 5: # ! 6: # double atof (s) ! 7: # char *s; ! 8: # ! 9: # Register usage: ! 10: # ! 11: # r0-1: value being developed ! 12: # r2: first section: pointer to the next character ! 13: # second section: binary exponent ! 14: # r3: flags ! 15: # r4: first section: the current character ! 16: # second section: scratch ! 17: # r5: the decimal exponent ! 18: # r6-7: scratch ! 19: # ! 20: # Flag definitions ! 21: # ! 22: .set msign,0 # mantissa has negative sign ! 23: .set esign,1 # exponent has negative sign ! 24: .set decpt,2 # decimal point encountered ! 25: ! 26: .align 2 ! 27: two31: .word 0x5000 # 2 ** 31 ! 28: .word 0 # (=2147483648) ! 29: .word 0 # in floating-point ! 30: .word 0 # (so atof doesn't have to convert it) ! 31: # ! 32: # Entry point ! 33: # ! 34: .text ! 35: .align 2 ! 36: .globl _atof ! 37: _atof: .word 0x00c0 # Save r7, r6 (we use r0-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: # ! 47: sk0: 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 ! 54: sk1: ! 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 $1<msign,r3 # Indicate a negative mantissa ! 63: cs1: movzbl (r2)+,r4 # Skip the character ! 64: cs2: ! 65: # ! 66: # Accumulate digits, keeping track of the exponent ! 67: # ! 68: clrq r0 # Clear the accumulator ! 69: ad0: cmpb r4,$'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: # ! 97: ad1: incl r5 # Ignore the digit, bump exponent ! 98: # ! 99: # If we have seen a decimal point, decrease the exponent by 1 ! 100: # ! 101: ad2: jbc $decpt,r3,ad3 # Jump if decimal point not seen ! 102: decl r5 # Decrease exponent ! 103: ad3: ! 104: # ! 105: # Fetch the next character, back for more ! 106: # ! 107: movzbl (r2)+,r4 # Fetch ! 108: jbr ad0 # Try again ! 109: # ! 110: # Not a digit. Could it be a decimal point? ! 111: # ! 112: ad4: cmpb r4,$'. # 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 $decpt,r3,ad3 # If it IS a decimal point, we record that ! 116: # we've seen one, and keep collecting ! 117: # digits if it is the first one. ! 118: # ! 119: # Check for an exponent ! 120: # ! 121: ad5: clrl r6 # Initialize the exponent accumulator ! 122: ! 123: cmpb r4,$'e # We allow both lower case e ! 124: jeql ex1 # ... and ... ! 125: cmpb r4,$'E # upper-case E ! 126: jneq ex7 ! 127: # ! 128: # Does the exponent have a sign? ! 129: # ! 130: ex1: movzbl (r2)+,r4 # Get next character ! 131: cmpb r4,$'+ # Positive sign? ! 132: jeql ex2 # ... yes ... ! 133: cmpb r4,$'- # Negative sign? ! 134: jneq ex3 # ... no ... ! 135: bisb2 $1<esign,r3 # Indicate exponent is negative ! 136: ex2: movzbl (r2)+,r4 # Grab the next character ! 137: # ! 138: # Accumulate exponent digits in r6 ! 139: # ! 140: ex3: cmpb r4,$'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 ! 149: ex4: ! 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 $esign,r3,ex5 # Jump if exponent not negative ! 157: mnegl r6,r6 # If sign, negate exponent ! 158: ex5: addl2 r6,r5 # Add given exponent to calculated exponent ! 159: cmpl r5,$-100 # Absurdly small? ! 160: jgtr ex6 # ... no ! 161: movl $-100,r5 # ... yes, force within limit ! 162: ex6: cmpl r5,$100 # Absurdly large? ! 163: jlss ex7 # ... no ! 164: movl $100,r5 # ... yes, force within bounds ! 165: ex7: ! 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: # ! 192: su0: 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: # ! 201: su1: 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: # ! 220: sd0: 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. ! 232: sd1: ashq $1,r0,r0 # Shift (r0,r1) left one, and ! 233: decl r2 # decrement the exponent to compensate ! 234: sd2: 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: # ! 239: sd3: 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: # ! 264: sd4: 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: # ! 285: cm0: 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: # ! 292: cm1: 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: # ! 315: cm2: pushl r2 # Put exponent in parameter list ! 316: movd r0,-(sp) # and also mantissa ! 317: calls $3,_ldexp # go combine them ! 318: ! 319: exit: ret
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