From: Catherine Moore Date: Mon, 29 Jun 1998 10:26:45 +0000 (+0000) Subject: config/sparc/lb1spc.asm (.udiv, .div) Replace routines. X-Git-Url: https://git.libre-soc.org/?a=commitdiff_plain;h=a611ae14b30a663ef25773902c3e40bc47acf17c;p=gcc.git config/sparc/lb1spc.asm (.udiv, .div) Replace routines. From-SVN: r20790 --- diff --git a/gcc/ChangeLog b/gcc/ChangeLog index fdf02373e5a..cb24b610c42 100644 --- a/gcc/ChangeLog +++ b/gcc/ChangeLog @@ -1,3 +1,7 @@ +Mon Jun 29 12:18:00 1998 Catherine Moore + + * config/lb1spc.asm (.div, .udiv): Replace routines. + Mon Jun 29 09:44:24 1998 Mark Mitchell * rtl.h: Update comment about special gen_rtx variants. diff --git a/gcc/config/sparc/lb1spc.asm b/gcc/config/sparc/lb1spc.asm index d4dea1ae765..831f33a988f 100644 --- a/gcc/config/sparc/lb1spc.asm +++ b/gcc/config/sparc/lb1spc.asm @@ -80,205 +80,361 @@ mul_shortway: #endif #ifdef L_divsi3 -.text - .align 4 - .global .udiv - .proc 4 +/* + * 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: - save %sp, -64, %sp - b divide - mov 0, %l2 ! result always positive - .global .div - .proc 4 + b ready_to_divide + mov 0, %g3 ! result is always positive + + .global .div + .align 4 + .proc 4 + .text .div: - save %sp, -64, %sp - orcc %i1, %i0, %g0 ! is either operand negative - bge divide ! if not, skip this junk - xor %i1, %i0, %l2 ! record sign of result in sign of %l2 - tst %i1 - bge 2f - tst %i0 - ! %i1 < 0 - bge divide - neg %i1 -2: ! %i0 < 0 - neg %i0 - ! FALL THROUGH -divide: - ! Compute size of quotient, scale comparand. - orcc %i1, %g0, %l1 ! movcc %i1, %l1 - te 2 ! if %i1 = 0 - mov %i0, %i3 - mov 0, %i2 - sethi %hi(1<<(32-4-1)), %l3 - cmp %i3, %l3 + ! 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 - mov 0, %l0 - ! - ! Here, the %i0 is >= 2^(31-3) or so. We must be careful here, - ! as our usual 3-at-a-shot divide step will cause overflow and havoc. - ! The total number of bits in the result here is 3*%l0+%l4, where - ! %l4 <= 3. - ! Compute %l0 in an unorthodox manner: know we need to Shift %l1 into - ! the top decade: so do not even bother to compare to %i3. -1: cmp %l1, %l3 - bgeu 3f - mov 1, %l4 - sll %l1, 3, %l1 - b 1b - inc %l0 - ! - ! Now compute %l4 - ! -2: addcc %l1, %l1, %l1 - bcc not_too_big - add %l4, 1, %l4 - ! - ! We are here if the %i1 overflowed when Shifting. - ! This means that %i3 has the high-order bit set. - ! Restore %l1 and subtract from %i3. - sll %l3, 4, %l3 - srl %l1, 1, %l1 - add %l1, %l3, %l1 - b do_single_div - dec %l4 -not_too_big: -3: cmp %l1, %i3 - blu 2b - nop - be do_single_div - nop - ! %l1 > %i3: went too far: back up 1 step - ! srl %l1, 1, %l1 - ! dec %l4 + 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 %i3 >= %l1, so we can do the + ! 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 %i3 >= 0. Because both %i3 and %l1 may have the - ! high-order bit set in the first step, just falling into the regular + ! 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: - deccc %l4 - bl end_regular_divide - nop - sub %i3, %l1, %i3 - mov 1, %i2 - b end_single_divloop - nop -single_divloop: - sll %i2, 1, %i2 - bl 1f - srl %l1, 1, %l1 - ! %i3 >= 0 - sub %i3, %l1, %i3 - b 2f - inc %i2 -1: ! %i3 < 0 - add %i3, %l1, %i3 - dec %i2 -end_single_divloop: -2: deccc %l4 - bge single_divloop - tst %i3 - b end_regular_divide - nop + 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 %l1, 3, %l1 - cmp %l1, %i3 +1: + sll %o5, 4, %o5 + cmp %o5, %o3 bleu 1b - inccc %l0 + addcc %o4, 1, %o4 be got_result - dec %l0 -do_regular_divide: - ! Do the main division iteration - tst %i3 - ! Fall through into divide loop + sub %o4, 1, %o4 + + tst %o3 ! set up for initial iteration divloop: - sll %i2, 3, %i2 + sll %o2, 4, %o2 ! depth 1, accumulated bits 0 - bl L.1.8 - srl %l1,1,%l1 + bl L1.16 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + subcc %o3,%o5,%o3 ! depth 2, accumulated bits 1 - bl L.2.9 - srl %l1,1,%l1 + bl L2.17 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + subcc %o3,%o5,%o3 ! depth 3, accumulated bits 3 - bl L.3.11 - srl %l1,1,%l1 + bl L3.19 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + 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 %i2, (3*2+1), %i2 -L.3.11: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (5*2+1), %o2 + +L4.21: + ! remainder is negative + addcc %o3,%o5,%o3 b 9f - add %i2, (3*2-1), %i2 -L.2.9: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (5*2-1), %o2 + +L2.17: + ! remainder is negative + addcc %o3,%o5,%o3 ! depth 3, accumulated bits 1 - bl L.3.9 - srl %l1,1,%l1 + bl L3.17 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + 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 %i2, (1*2+1), %i2 -L.3.9: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (3*2+1), %o2 + +L4.19: + ! remainder is negative + addcc %o3,%o5,%o3 b 9f - add %i2, (1*2-1), %i2 -L.1.8: ! remainder is negative - addcc %i3,%l1,%i3 + 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 L.2.7 - srl %l1,1,%l1 + bl L2.15 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + subcc %o3,%o5,%o3 ! depth 3, accumulated bits -1 - bl L.3.7 - srl %l1,1,%l1 + 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 %i3,%l1,%i3 + subcc %o3,%o5,%o3 b 9f - add %i2, (-1*2+1), %i2 -L.3.7: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (-3*2+1), %o2 + +L4.13: + ! remainder is negative + addcc %o3,%o5,%o3 b 9f - add %i2, (-1*2-1), %i2 -L.2.7: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (-3*2-1), %o2 + +L2.15: + ! remainder is negative + addcc %o3,%o5,%o3 ! depth 3, accumulated bits -3 - bl L.3.5 - srl %l1,1,%l1 + bl L3.13 + srl %o5,1,%o5 ! remainder is positive - subcc %i3,%l1,%i3 + 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 %i2, (-3*2+1), %i2 -L.3.5: ! remainder is negative - addcc %i3,%l1,%i3 + add %o2, (-5*2+1), %o2 + +L4.11: + ! remainder is negative + addcc %o3,%o5,%o3 b 9f - add %i2, (-3*2-1), %i2 + 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: -9: deccc %l0 + subcc %o4, 1, %o4 bge divloop - tst %i3 - bge got_result - nop - ! non-restoring fixup here - dec %i2 + tst %o3 + bl,a got_result + ! non-restoring fixup here (one instruction only!) + sub %o2, 1, %o2 + + got_result: - tst %l2 - bge 1f - restore - ! answer < 0 - retl ! leaf-routine return - neg %o2, %o0 ! quotient <- -%i2 -1: retl ! leaf-routine return - mov %o2, %o0 ! quotient <- %i2 + ! 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