| /* This is an assembly language implementation of mulsi3, divsi3, and modsi3 |
| for the sparc processor. |
| |
| These routines are derived from the SPARC Architecture Manual, version 8, |
| slightly edited to match the desired calling convention, and also to |
| optimize them for our purposes. */ |
| |
| /* An executable stack is *not* required for these functions. */ |
| #if defined(__ELF__) && defined(__linux__) |
| .section .note.GNU-stack,"",%progbits |
| .previous |
| #endif |
| |
| #ifdef L_mulsi3 |
| .text |
| .align 4 |
| .global .umul |
| .proc 4 |
| .umul: |
| or %o0, %o1, %o4 ! logical or of multiplier and multiplicand |
| mov %o0, %y ! multiplier to Y register |
| andncc %o4, 0xfff, %o5 ! mask out lower 12 bits |
| be mul_shortway ! can do it the short way |
| andcc %g0, %g0, %o4 ! zero the partial product and clear NV cc |
| ! |
| ! long multiply |
| ! |
| mulscc %o4, %o1, %o4 ! first iteration of 33 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 ! 32nd iteration |
| mulscc %o4, %g0, %o4 ! last iteration only shifts |
| ! the upper 32 bits of product are wrong, but we do not care |
| retl |
| rd %y, %o0 |
| ! |
| ! short multiply |
| ! |
| mul_shortway: |
| mulscc %o4, %o1, %o4 ! first iteration of 13 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 |
| mulscc %o4, %o1, %o4 ! 12th iteration |
| mulscc %o4, %g0, %o4 ! last iteration only shifts |
| rd %y, %o5 |
| sll %o4, 12, %o4 ! left shift partial product by 12 bits |
| srl %o5, 20, %o5 ! right shift partial product by 20 bits |
| retl |
| or %o5, %o4, %o0 ! merge for true product |
| #endif |
| |
| #ifdef L_divsi3 |
| /* |
| * 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: |
| * .div name of function to generate |
| * div div=div => %o0 / %o1; div=rem => %o0 % %o1 |
| * true true=true => signed; true=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. |
| */ |
| .global .udiv |
| .align 4 |
| .proc 4 |
| .text |
| .udiv: |
| b ready_to_divide |
| mov 0, %g3 ! result is always positive |
| |
| .global .div |
| .align 4 |
| .proc 4 |
| .text |
| .div: |
| ! compute sign of result; if neither is negative, no problem |
| orcc %o1, %o0, %g0 ! either negative? |
| bge ready_to_divide ! no, go do the divide |
| xor %o1, %o0, %g3 ! compute sign in any case |
| tst %o1 |
| bge 1f |
| tst %o0 |
| ! %o1 is definitely negative; %o0 might also be negative |
| bge ready_to_divide ! if %o0 not negative... |
| sub %g0, %o1, %o1 ! in any case, make %o1 nonneg |
| 1: ! %o0 is negative, %o1 is nonnegative |
| sub %g0, %o0, %o0 ! make %o0 nonnegative |
| |
| |
| ready_to_divide: |
| |
| ! 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 0x2 ! ST_DIV0 |
| retl |
| clr %o0 |
| |
| 1: |
| cmp %o3, %o5 ! if %o1 exceeds %o0, done |
| blu got_result ! (and algorithm fails otherwise) |
| clr %o2 |
| sethi %hi(1 << (32 - 4 - 1)), %g1 |
| cmp %o3, %g1 |
| blu not_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, %g2 |
| sll %o5, 4, %o5 |
| b 1b |
| add %o4, 1, %o4 |
| |
| ! Now compute %g2. |
| 2: addcc %o5, %o5, %o5 |
| bcc not_too_big |
| add %g2, 1, %g2 |
| |
| ! 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 do_single_div |
| sub %g2, 1, %g2 |
| |
| not_too_big: |
| 3: cmp %o5, %o3 |
| blu 2b |
| nop |
| be do_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 %g2 |
| ! 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... |
| do_single_div: |
| subcc %g2, 1, %g2 |
| bl end_regular_divide |
| nop |
| sub %o3, %o5, %o3 |
| mov 1, %o2 |
| b end_single_divloop |
| nop |
| single_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: |
| end_single_divloop: |
| subcc %g2, 1, %g2 |
| bge single_divloop |
| tst %o3 |
| b,a end_regular_divide |
| |
| not_really_big: |
| 1: |
| sll %o5, 4, %o5 |
| cmp %o5, %o3 |
| bleu 1b |
| addcc %o4, 1, %o4 |
| be got_result |
| sub %o4, 1, %o4 |
| |
| tst %o3 ! set up for initial iteration |
| divloop: |
| sll %o2, 4, %o2 |
| ! depth 1, accumulated bits 0 |
| bl L1.16 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 2, accumulated bits 1 |
| bl L2.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits 3 |
| bl L3.19 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 7 |
| bl L4.23 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (7*2+1), %o2 |
| |
| L4.23: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (7*2-1), %o2 |
| |
| |
| L3.19: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 5 |
| bl L4.21 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (5*2+1), %o2 |
| |
| L4.21: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (5*2-1), %o2 |
| |
| L2.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits 1 |
| bl L3.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 3 |
| bl L4.19 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (3*2+1), %o2 |
| |
| L4.19: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (3*2-1), %o2 |
| |
| L3.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 1 |
| bl L4.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (1*2+1), %o2 |
| |
| L4.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (1*2-1), %o2 |
| |
| L1.16: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 2, accumulated bits -1 |
| bl L2.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits -1 |
| bl L3.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -1 |
| bl L4.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-1*2+1), %o2 |
| |
| L4.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-1*2-1), %o2 |
| |
| L3.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -3 |
| bl L4.13 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-3*2+1), %o2 |
| |
| L4.13: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-3*2-1), %o2 |
| |
| L2.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits -3 |
| bl L3.13 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -5 |
| bl L4.11 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-5*2+1), %o2 |
| |
| L4.11: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-5*2-1), %o2 |
| |
| L3.13: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -7 |
| bl L4.9 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-7*2+1), %o2 |
| |
| L4.9: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-7*2-1), %o2 |
| |
| 9: |
| end_regular_divide: |
| subcc %o4, 1, %o4 |
| bge divloop |
| tst %o3 |
| bl,a got_result |
| ! non-restoring fixup here (one instruction only!) |
| sub %o2, 1, %o2 |
| |
| |
| got_result: |
| ! check to see if answer should be < 0 |
| tst %g3 |
| bl,a 1f |
| sub %g0, %o2, %o2 |
| 1: |
| retl |
| mov %o2, %o0 |
| #endif |
| |
| #ifdef L_modsi3 |
| /* This implementation was taken from glibc: |
| * |
| * Input: dividend and divisor in %o0 and %o1 respectively. |
| * |
| * 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. |
| */ |
| .text |
| .align 4 |
| .global .urem |
| .proc 4 |
| .urem: |
| b divide |
| mov 0, %g3 ! result always positive |
| |
| .align 4 |
| .global .rem |
| .proc 4 |
| .rem: |
| ! compute sign of result; if neither is negative, no problem |
| orcc %o1, %o0, %g0 ! either negative? |
| bge 2f ! no, go do the divide |
| mov %o0, %g3 ! sign of remainder matches %o0 |
| tst %o1 |
| bge 1f |
| tst %o0 |
| ! %o1 is definitely negative; %o0 might also be negative |
| bge 2f ! if %o0 not negative... |
| sub %g0, %o1, %o1 ! in any case, make %o1 nonneg |
| 1: ! %o0 is negative, %o1 is nonnegative |
| sub %g0, %o0, %o0 ! make %o0 nonnegative |
| 2: |
| |
| ! Ready to divide. Compute size of quotient; scale comparand. |
| divide: |
| 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 0x2 !ST_DIV0 |
| retl |
| clr %o0 |
| |
| 1: |
| cmp %o3, %o5 ! if %o1 exceeds %o0, done |
| blu got_result ! (and algorithm fails otherwise) |
| clr %o2 |
| sethi %hi(1 << (32 - 4 - 1)), %g1 |
| cmp %o3, %g1 |
| blu not_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, %g2 |
| sll %o5, 4, %o5 |
| b 1b |
| add %o4, 1, %o4 |
| |
| ! Now compute %g2. |
| 2: addcc %o5, %o5, %o5 |
| bcc not_too_big |
| add %g2, 1, %g2 |
| |
| ! 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 do_single_div |
| sub %g2, 1, %g2 |
| |
| not_too_big: |
| 3: cmp %o5, %o3 |
| blu 2b |
| nop |
| be do_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 %g2 |
| ! 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... |
| do_single_div: |
| subcc %g2, 1, %g2 |
| bl end_regular_divide |
| nop |
| sub %o3, %o5, %o3 |
| mov 1, %o2 |
| b end_single_divloop |
| nop |
| single_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: |
| end_single_divloop: |
| subcc %g2, 1, %g2 |
| bge single_divloop |
| tst %o3 |
| b,a end_regular_divide |
| |
| not_really_big: |
| 1: |
| sll %o5, 4, %o5 |
| cmp %o5, %o3 |
| bleu 1b |
| addcc %o4, 1, %o4 |
| be got_result |
| sub %o4, 1, %o4 |
| |
| tst %o3 ! set up for initial iteration |
| divloop: |
| sll %o2, 4, %o2 |
| ! depth 1, accumulated bits 0 |
| bl L1.16 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 2, accumulated bits 1 |
| bl L2.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits 3 |
| bl L3.19 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 7 |
| bl L4.23 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (7*2+1), %o2 |
| L4.23: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (7*2-1), %o2 |
| |
| L3.19: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 5 |
| bl L4.21 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (5*2+1), %o2 |
| |
| L4.21: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (5*2-1), %o2 |
| |
| L2.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits 1 |
| bl L3.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 3 |
| bl L4.19 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (3*2+1), %o2 |
| |
| L4.19: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (3*2-1), %o2 |
| |
| L3.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits 1 |
| bl L4.17 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (1*2+1), %o2 |
| |
| L4.17: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (1*2-1), %o2 |
| |
| L1.16: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 2, accumulated bits -1 |
| bl L2.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits -1 |
| bl L3.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -1 |
| bl L4.15 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-1*2+1), %o2 |
| |
| L4.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-1*2-1), %o2 |
| |
| L3.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -3 |
| bl L4.13 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-3*2+1), %o2 |
| |
| L4.13: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-3*2-1), %o2 |
| |
| L2.15: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 3, accumulated bits -3 |
| bl L3.13 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -5 |
| bl L4.11 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-5*2+1), %o2 |
| |
| L4.11: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-5*2-1), %o2 |
| |
| L3.13: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| ! depth 4, accumulated bits -7 |
| bl L4.9 |
| srl %o5,1,%o5 |
| ! remainder is positive |
| subcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-7*2+1), %o2 |
| |
| L4.9: |
| ! remainder is negative |
| addcc %o3,%o5,%o3 |
| b 9f |
| add %o2, (-7*2-1), %o2 |
| |
| 9: |
| end_regular_divide: |
| subcc %o4, 1, %o4 |
| bge divloop |
| tst %o3 |
| bl,a got_result |
| ! non-restoring fixup here (one instruction only!) |
| add %o3, %o1, %o3 |
| |
| got_result: |
| ! check to see if answer should be < 0 |
| tst %g3 |
| bl,a 1f |
| sub %g0, %o3, %o3 |
| 1: |
| retl |
| mov %o3, %o0 |
| |
| #endif |
| |