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| Current File : //usr/src/contrib/compiler-rt/lib/sparc64/modsi3.S |
/* * This m4 code has been taken from The SPARC Architecture Manual Version 8. */ /* * Division/Remainder * * Input is: * dividend -- the thing being divided * divisor -- how many ways to divide it * Important parameters: * N -- how many bits per iteration we try to get * as our current guess: * WORDSIZE -- how many bits altogether we're talking about: * obviously: * A derived constant: * TOPBITS -- how many bits are in the top "decade" of a number: * * Important variables are: * Q -- the partial quotient under development -- initially 0 * R -- the remainder so far -- initially == the dividend * ITER -- number of iterations of the main division loop which will * be required. Equal to CEIL( lg2(quotient)/4 ) * Note that this is log_base_(2ˆ4) of the quotient. * V -- the current comparand -- initially divisor*2ˆ(ITER*4-1) * Cost: * current estimate for non-large dividend is * CEIL( lg2(quotient) / 4 ) x ( 10 + 74/2 ) + C * a large dividend is one greater than 2ˆ(31-4 ) and takes a * different path, as the upper bits of the quotient must be developed * one bit at a time. * This uses the m4 and cpp macro preprocessors. */ /* * This is the recursive definition of how we develop quotient digits. * It takes three important parameters: * $1 -- the current depth, 1<=$1<=4 * $2 -- the current accumulation of quotient bits * 4 -- max depth * We add a new bit to $2 and either recurse or insert the bits in the quotient. * Dynamic input: * %o3 -- current remainder * %o2 -- current quotient * %o5 -- current comparand * cc -- set on current value of %o3 * Dynamic output: * %o3', %o2', %o5', cc' */ #include "../assembly.h" .text .align 32 DEFINE_COMPILERRT_FUNCTION(__umodsi3) b divide mov 0,%g3 ! result always nonnegative .text .align 32 DEFINE_COMPILERRT_FUNCTION(__modsi3) orcc %o1,%o0,%g0 ! are either %o0 or %o1 negative bge divide ! if not, skip this junk mov %o0,%g3 ! record sign of result in sign of %g3 tst %o1 bge 2f tst %o0 ! %o1 < 0 bge divide neg %o1 2: ! %o0 < 0 neg %o0 ! FALL THROUGH divide: ! Compute size of quotient, scale comparand. orcc %o1,%g0,%o5 ! movcc %o1,%o5 te 2 ! if %o1 = 0 mov %o0,%o3 mov 0,%o2 sethi %hi(1<<(32-4 -1)),%g1 cmp %o3,%g1 blu not_really_big mov 0,%o4 ! ! Here, the %o0 is >= 2ˆ(31-4) or so. We must be careful here, ! as our usual 4-at-a-shot divide step will cause overflow and havoc. ! The total number of bits in the result here is 4*%o4+%g2, where ! %g2 <= 4. ! Compute %o4 in an unorthodox manner: know we need to Shift %o5 into ! the top decade: so don't even bother to compare to %o3. 1: cmp %o5,%g1 bgeu 3f mov 1,%g2 sll %o5,4,%o5 b 1b inc %o4 ! Now compute %g2 2: addcc %o5,%o5,%o5 bcc not_too_big add %g2,1,%g2 ! We're here if the %o1 overflowed when 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 dec %g2 not_too_big: 3: cmp %o5,%o3 blu 2b nop be do_single_div nop ! %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: deccc %g2 bl end_regular_divide nop sub %o3,%o5,%o3 mov 1,%o2 b,a end_single_divloop ! EMPTY single_divloop: sll %o2,1,%o2 bl 1f srl %o5,1,%o5 ! %o3 >= 0 sub %o3,%o5,%o3 b 2f inc %o2 1: ! %o3 < 0 add %o3,%o5,%o3 dec %o2 2: end_single_divloop: deccc %g2 bge single_divloop tst %o3 b,a end_regular_divide ! EMPTY not_really_big: 1: sll %o5,4,%o5 cmp %o5,%o3 bleu 1b inccc %o4 be got_result dec %o4 do_regular_divide: ! Do the main division iteration tst %o3 ! Fall through into divide loop divloop: sll %o2,4,%o2 !depth 1, accumulated bits 0 bl L.1.16 srl %o5,1,%o5 ! remainder is nonnegative subcc %o3,%o5,%o3 !depth 2, accumulated bits 1 bl L.2.17 srl %o5,1,%o5 ! remainder is nonnegative subcc %o3,%o5,%o3 !depth 3, accumulated bits 3 bl L.3.19 srl %o5,1,%o5 ! remainder is nonnegative subcc %o3,%o5,%o3 !depth 4, accumulated bits 7 bl L.4.23 srl %o5,1,%o5 ! remainder is nonnegative 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 nonnegative 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 nonnegative subcc %o3,%o5,%o3 !depth 4, accumulated bits 3 bl L.4.19 srl %o5,1,%o5 ! remainder is nonnegative 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 nonnegative 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 nonnegative subcc %o3,%o5,%o3 !depth 3, accumulated bits -1 bl L.3.15 srl %o5,1,%o5 ! remainder is nonnegative subcc %o3,%o5,%o3 !depth 4, accumulated bits -1 bl L.4.15 srl %o5,1,%o5 ! remainder is nonnegative 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 nonnegative 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 nonnegative subcc %o3,%o5,%o3 !depth 4, accumulated bits -5 bl L.4.11 srl %o5,1,%o5 ! remainder is nonnegative 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 nonnegative 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: end_regular_divide: deccc %o4 bge divloop tst %o3 bl,a got_result ! non-restoring fixup if remainder < 0, otherwise annulled add %o3,%o1,%o3 got_result: tst %g3 bl,a 1f ! negate for answer < 0, otherwise annulled neg %o3,%o3 ! %o3 <- -%o3 1: retl ! leaf-routine return mov %o3,%o0 ! remainder <- %o3