diff options
Diffstat (limited to 'gcc-4.9/libgcc/config/sparc/lb1spc.S')
| -rw-r--r-- | gcc-4.9/libgcc/config/sparc/lb1spc.S | 784 |
1 files changed, 784 insertions, 0 deletions
diff --git a/gcc-4.9/libgcc/config/sparc/lb1spc.S b/gcc-4.9/libgcc/config/sparc/lb1spc.S new file mode 100644 index 000000000..b60bd5740 --- /dev/null +++ b/gcc-4.9/libgcc/config/sparc/lb1spc.S @@ -0,0 +1,784 @@ +/* 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. */ + +#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 + |