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/* $Id: rem.S,v 1.1 2003/04/14 03:24:43 bencollins Exp $
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* rem.S: This routine was taken from glibc-1.09 and is covered
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* by the GNU Library General Public License Version 2.
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/* This file is generated from divrem.m4; DO NOT EDIT! */
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* Division and remainder, from Appendix E of the Sparc Version 8
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* Architecture Manual, with fixes from Gordon Irlam.
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* Input: dividend and divisor in %o0 and %o1 respectively.
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* .rem name of function to generate
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* rem rem=div => %o0 / %o1; rem=rem => %o0 % %o1
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* true true=true => signed; true=false => unsigned
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* Algorithm parameters:
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* N how many bits per iteration we try to get (4)
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* WORDSIZE total number of bits (32)
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* TOPBITS number of bits in the top decade of a number
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* Important variables:
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* Q the partial quotient under development (initially 0)
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* R the remainder so far, initially the dividend
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* ITER number of main division loop iterations required;
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* equal to ceil(log2(quotient) / N). Note that this
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* is the log base (2^N) of the quotient.
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* V the current comparand, initially divisor*2^(ITER*N-1)
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* Current estimate for non-large dividend is
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* ceil(log2(quotient) / N) * (10 + 7N/2) + C
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* A large dividend is one greater than 2^(31-TOPBITS) and takes a
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* different path, as the upper bits of the quotient must be developed
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.register %g2,#scratch
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! compute sign of result; if neither is negative, no problem
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orcc %o1, %o0, %g0 ! either negative?
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bge 2f ! no, go do the divide
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mov %o0, %g2 ! compute sign in any case
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! %o1 is definitely negative; %o0 might also be negative
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bge 2f ! if %o0 not negative...
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sub %g0, %o1, %o1 ! in any case, make %o1 nonneg
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1: ! %o0 is negative, %o1 is nonnegative
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sub %g0, %o0, %o0 ! make %o0 nonnegative
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! Ready to divide. Compute size of quotient; scale comparand.
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! Divide by zero trap. If it returns, return 0 (about as
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! wrong as possible, but that is what SunOS does...).
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! ta ST_DIV0 -- Not for SILO
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cmp %o3, %o5 ! if %o1 exceeds %o0, done
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blu Lgot_result ! (and algorithm fails otherwise)
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sethi %hi(1 << (32 - 4 - 1)), %g1
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! Here the dividend is >= 2**(31-N) or so. We must be careful here,
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! as our usual N-at-a-shot divide step will cause overflow and havoc.
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! The number of bits in the result here is N*ITER+SC, where SC <= N.
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! Compute ITER in an unorthodox manner: know we need to shift V into
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! the top decade: so do not even bother to compare to R.
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! We get here if the %o1 overflowed while shifting.
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! This means that %o3 has the high-order bit set.
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! Restore %o5 and subtract from %o3.
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sll %g1, 4, %g1 ! high order bit
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srl %o5, 1, %o5 ! rest of %o5
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/* NB: these are commented out in the V8-Sparc manual as well */
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/* (I do not understand this) */
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! %o5 > %o3: went too far: back up 1 step
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! do single-bit divide steps
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! We have to be careful here. We know that %o3 >= %o5, so we can do the
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! first divide step without thinking. BUT, the others are conditional,
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! and are only done if %o3 >= 0. Because both %o3 and %o5 may have the high-
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! order bit set in the first step, just falling into the regular
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! division loop will mess up the first time around.
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! So we unroll slightly...
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bl Lend_regular_divide
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b Lend_single_divloop
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b,a Lend_regular_divide
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tst %o3 ! set up for initial iteration
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! depth 1, accumulated bits 0
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! remainder is positive
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! depth 2, accumulated bits 1
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! remainder is positive
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! depth 3, accumulated bits 3
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! remainder is positive
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! depth 4, accumulated bits 7
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! remainder is positive
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add %o2, (7*2+1), %o2
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! remainder is negative
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add %o2, (7*2-1), %o2
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! remainder is negative
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! depth 4, accumulated bits 5
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! remainder is positive
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add %o2, (5*2+1), %o2
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! remainder is negative
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add %o2, (5*2-1), %o2
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! remainder is negative
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! depth 3, accumulated bits 1
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! remainder is positive
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! depth 4, accumulated bits 3
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! remainder is positive
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add %o2, (3*2+1), %o2
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! remainder is negative
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add %o2, (3*2-1), %o2
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! remainder is negative
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! depth 4, accumulated bits 1
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! remainder is positive
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add %o2, (1*2+1), %o2
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! remainder is negative
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add %o2, (1*2-1), %o2
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! remainder is negative
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! depth 2, accumulated bits -1
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! remainder is positive
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! depth 3, accumulated bits -1
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! remainder is positive
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! depth 4, accumulated bits -1
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! remainder is positive
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add %o2, (-1*2+1), %o2
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! remainder is negative
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add %o2, (-1*2-1), %o2
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! remainder is negative
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! depth 4, accumulated bits -3
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! remainder is positive
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add %o2, (-3*2+1), %o2
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! remainder is negative
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add %o2, (-3*2-1), %o2
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! remainder is negative
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! depth 3, accumulated bits -3
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! remainder is positive
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! depth 4, accumulated bits -5
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! remainder is positive
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add %o2, (-5*2+1), %o2
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! remainder is negative
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add %o2, (-5*2-1), %o2
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! remainder is negative
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! depth 4, accumulated bits -7
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! remainder is positive
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add %o2, (-7*2+1), %o2
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! remainder is negative
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add %o2, (-7*2-1), %o2
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! non-restoring fixup here (one instruction only!)
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! check to see if answer should be < 0
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1: smul %o2, %o1, %o2
added
removed
common/rem.S
