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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 | /* $Id: urem.S,v 1.4 1996/09/30 02:22:42 davem Exp $ * urem.S: This routine was taken from glibc-1.09 and is covered * by the GNU Library General Public License Version 2. */ /* This file is generated from divrem.m4; DO NOT EDIT! */ /* * Division and remainder, from Appendix E of the Sparc Version 8 * Architecture Manual, with fixes from Gordon Irlam. */ /* * Input: dividend and divisor in %o0 and %o1 respectively. * * m4 parameters: * .urem name of function to generate * rem rem=div => %o0 / %o1; rem=rem => %o0 % %o1 * false false=true => signed; false=false => unsigned * * Algorithm parameters: * N how many bits per iteration we try to get (4) * WORDSIZE total number of bits (32) * * Derived constants: * TOPBITS number of bits in the top decade of a number * * Important variables: * Q the partial quotient under development (initially 0) * R the remainder so far, initially the dividend * ITER number of main division loop iterations required; * equal to ceil(log2(quotient) / N). Note that this * is the log base (2^N) of the quotient. * V the current comparand, initially divisor*2^(ITER*N-1) * * Cost: * Current estimate for non-large dividend is * ceil(log2(quotient) / N) * (10 + 7N/2) + C * A large dividend is one greater than 2^(31-TOPBITS) and takes a * different path, as the upper bits of the quotient must be developed * one bit at a time. */ .globl .urem .urem: ! Ready to divide. Compute size of quotient; scale comparand. orcc %o1, %g0, %o5 bne 1f mov %o0, %o3 ! Divide by zero trap. If it returns, return 0 (about as ! wrong as possible, but that is what SunOS does...). ta ST_DIV0 retl clr %o0 1: cmp %o3, %o5 ! if %o1 exceeds %o0, done blu Lgot_result ! (and algorithm fails otherwise) clr %o2 sethi %hi(1 << (32 - 4 - 1)), %g1 cmp %o3, %g1 blu Lnot_really_big clr %o4 ! Here the dividend is >= 2**(31-N) or so. We must be careful here, ! as our usual N-at-a-shot divide step will cause overflow and havoc. ! The number of bits in the result here is N*ITER+SC, where SC <= N. ! Compute ITER in an unorthodox manner: know we need to shift V into ! the top decade: so do not even bother to compare to R. 1: cmp %o5, %g1 bgeu 3f mov 1, %g7 sll %o5, 4, %o5 b 1b add %o4, 1, %o4 ! Now compute %g7. 2: addcc %o5, %o5, %o5 bcc Lnot_too_big add %g7, 1, %g7 ! We get here if the %o1 overflowed while shifting. ! This means that %o3 has the high-order bit set. ! Restore %o5 and subtract from %o3. sll %g1, 4, %g1 ! high order bit srl %o5, 1, %o5 ! rest of %o5 add %o5, %g1, %o5 b Ldo_single_div sub %g7, 1, %g7 Lnot_too_big: 3: cmp %o5, %o3 blu 2b nop be Ldo_single_div nop /* NB: these are commented out in the V8-Sparc manual as well */ /* (I do not understand this) */ ! %o5 > %o3: went too far: back up 1 step ! srl %o5, 1, %o5 ! dec %g7 ! do single-bit divide steps ! ! We have to be careful here. We know that %o3 >= %o5, so we can do the ! first divide step without thinking. BUT, the others are conditional, ! and are only done if %o3 >= 0. Because both %o3 and %o5 may have the high- ! order bit set in the first step, just falling into the regular ! division loop will mess up the first time around. ! So we unroll slightly... Ldo_single_div: subcc %g7, 1, %g7 bl Lend_regular_divide nop sub %o3, %o5, %o3 mov 1, %o2 b Lend_single_divloop nop Lsingle_divloop: sll %o2, 1, %o2 bl 1f srl %o5, 1, %o5 ! %o3 >= 0 sub %o3, %o5, %o3 b 2f add %o2, 1, %o2 1: ! %o3 < 0 add %o3, %o5, %o3 sub %o2, 1, %o2 2: Lend_single_divloop: subcc %g7, 1, %g7 bge Lsingle_divloop tst %o3 b,a Lend_regular_divide Lnot_really_big: 1: sll %o5, 4, %o5 cmp %o5, %o3 bleu 1b addcc %o4, 1, %o4 be Lgot_result sub %o4, 1, %o4 tst %o3 ! set up for initial iteration Ldivloop: sll %o2, 4, %o2 ! depth 1, accumulated bits 0 bl L.1.16 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 2, accumulated bits 1 bl L.2.17 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 3, accumulated bits 3 bl L.3.19 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 4, accumulated bits 7 bl L.4.23 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (7*2+1), %o2 L.4.23: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (7*2-1), %o2 L.3.19: ! remainder is negative addcc %o3,%o5,%o3 ! depth 4, accumulated bits 5 bl L.4.21 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (5*2+1), %o2 L.4.21: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (5*2-1), %o2 L.2.17: ! remainder is negative addcc %o3,%o5,%o3 ! depth 3, accumulated bits 1 bl L.3.17 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 4, accumulated bits 3 bl L.4.19 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (3*2+1), %o2 L.4.19: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (3*2-1), %o2 L.3.17: ! remainder is negative addcc %o3,%o5,%o3 ! depth 4, accumulated bits 1 bl L.4.17 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (1*2+1), %o2 L.4.17: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (1*2-1), %o2 L.1.16: ! remainder is negative addcc %o3,%o5,%o3 ! depth 2, accumulated bits -1 bl L.2.15 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 3, accumulated bits -1 bl L.3.15 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 4, accumulated bits -1 bl L.4.15 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (-1*2+1), %o2 L.4.15: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (-1*2-1), %o2 L.3.15: ! remainder is negative addcc %o3,%o5,%o3 ! depth 4, accumulated bits -3 bl L.4.13 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (-3*2+1), %o2 L.4.13: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (-3*2-1), %o2 L.2.15: ! remainder is negative addcc %o3,%o5,%o3 ! depth 3, accumulated bits -3 bl L.3.13 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 ! depth 4, accumulated bits -5 bl L.4.11 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (-5*2+1), %o2 L.4.11: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (-5*2-1), %o2 L.3.13: ! remainder is negative addcc %o3,%o5,%o3 ! depth 4, accumulated bits -7 bl L.4.9 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (-7*2+1), %o2 L.4.9: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (-7*2-1), %o2 9: Lend_regular_divide: subcc %o4, 1, %o4 bge Ldivloop tst %o3 bl,a Lgot_result ! non-restoring fixup here (one instruction only!) add %o3, %o1, %o3 Lgot_result: retl mov %o3, %o0 .globl .urem_patch .urem_patch: wr %g0, 0x0, %y nop nop nop udiv %o0, %o1, %o2 umul %o2, %o1, %o2 retl sub %o0, %o2, %o0 |