📄 sadiv.s
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/* * .seg "data" * .asciz "@(#)sadiv.s 1.1 92/07/30" * Copyright (c) 1986 by Sun Microsystems, Inc. */ .seg "text" .align 4/* * divison/remainder * * Input is: * dividend -- the thing being divided * divisor -- how many ways to divide * 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 -- initally 0 * R -- the remainder so far -- initially == the dividend * ITER -- number of iterations of the main division loop will * be required. Equal to CEIL( lg2(quotient)/2 ) * Note that this is log_base_(2^2) of the quotient. * V -- the current comparand -- initially divisor*2^(ITER*2-1) * Cost: * current estimate for non-large dividend is * CEIL( lg2(quotient) / 2 ) x ( 10 + 72/2 ) + C * a large dividend is one greater than 2^(31-2 ) and takes a * different path, as the upper bits of the quotient must be developed * one bit at a time. */#include <machine/trap.h>#include <machine/asm_linkage.h>/* * this is the recursive definition of how we develop quotient digits. * it takes three important parameters: * $1 -- the current depth, 1<=$1<=2 * $2 -- the current accumulation of quotient bits * 2 -- 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' */ RTENTRY(.rem) mov 1,%g4 ! signify remainder orcc %o1,%o0,%g0 ! are either %o0 or %o1 negative bge divide mov %o0, %g1 ! record sign of result in sign of %g1 b,a fixops RTENTRY(.urem) mov 1,%g4 ! signify remainder b divide ! next instruction needed in delay slot .empty ! keep assembler from complaining RTENTRY(.udiv) ! UNSIGNED DIVIDE mov 0, %g1 ! result always positive b divide ! next instruction needed in delay slot .empty ! keep assembler from complaining RTENTRY(.div) ! SIGNED DIVIDE mov 0, %g4 ! signify divide orcc %o1,%o0,%g0 ! are either %o0 or %o1 negative bge divide ! if not, skip this junk xor %o1,%o0,%g1 ! record sign of result in sign of %g1fixops: tst %o1 bge 2f tst %o0 ! %o1 < 0 bge divide neg %o1 2: ! %o0 < 0 neg %o0 ! FALL THROUGHdivide:! compute size of quotient, scale comparand orcc %o1,%g0,%o5 ! movcc %o1,%o5 te ST_DIV0 ! if %o1 = 0 mov %o0,%o3 cmp %o3,%o5 blu got_result ! if %o3<%o5 already, there's no point in continuing mov 0,%o2 sethi %hi(1<<(32-2 -1)),%g2 cmp %o3,%g2 blu not_really_big mov 0,%o4 ! ! here, the %o0 is >= 2^(31-2) or so. We must be careful here, as ! our usual 2-at-a-shot divide step will cause overflow and havoc. The ! total number of bits in the result here is 2*%o4+%g3, where %g3 <= 2. ! 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,%g2 bgeu 3f mov 1,%g3 sll %o5,2,%o5 b 1b inc %o4 ! now compute %g3 2: addcc %o5,%o5,%o5 bcc not_too_big ! bcc not_too_big add %g3,1,%g3 ! ! here if the %o1 overflowed when Shifting ! this means that %o3 has the high-order bit set ! restore %o5 and subtract from %o3 sll %g2,2 ,%g2 ! high order bit srl %o5,1,%o5 ! rest of %o5 add %o5,%g2,%o5 b do_single_div sub %g3,1,%g3 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 %g3 ! 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 %g3 bl end_regular_divide nop sub %o3,%o5,%o3 mov 1,%o2 b,a end_single_divloop 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 %g3 bge single_divloop tst %o3 b,a end_regular_dividenot_really_big:1: sll %o5,2,%o5 cmp %o5,%o3 bleu 1b inccc %o4 be got_result dec %o4do_regular_divide:! do the main division iteration tst %o3! fall through into divide loopdivloop: sll %o2,2,%o2 !depth 1, accumulated bits 0 bl L.1.4 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 !depth 2, accumulated bits 1 bl L.2.5 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (1*2+1), %o2 L.2.5: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (1*2-1), %o2 L.1.4: ! remainder is negative addcc %o3,%o5,%o3 !depth 2, accumulated bits -1 bl L.2.3 srl %o5,1,%o5 ! remainder is positive subcc %o3,%o5,%o3 b 9f add %o2, (-1*2+1), %o2 L.2.3: ! remainder is negative addcc %o3,%o5,%o3 b 9f add %o2, (-1*2-1), %o29:end_regular_divide: deccc %o4 bge divloop tst %o3 tst %g4 bz divout tst %o3 bl,a got_result add %o3,%o1,%o3 b,a remresultdivout: bl,a got_result dec %o2got_result: tst %g4 bz divresult .empty ! keep assembler from complainingremresult: tst %g1 bl,a 1f neg %o3 ! remainder <- -%o31: retl mov %o3,%o0 ! remainder <- %o3divresult: bl,a 1f neg %o2 ! quotient <- -%o21: retl mov %o2,%o0 ! quotient <- %o2
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