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235 lines
5.8 KiB
Plaintext
235 lines
5.8 KiB
Plaintext
/*
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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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*/
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/*
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* Input: dividend and divisor in %o0 and %o1 respectively.
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*
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* m4 parameters:
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* NAME name of function to generate
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* OP OP=div => %o0 / %o1; OP=rem => %o0 % %o1
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* S S=true => signed; S=false => unsigned
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*
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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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*
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* Derived constants:
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* TOPBITS number of bits in the top `decade' of a number
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*
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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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*
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* Cost:
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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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* one bit at a time.
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*/
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define(N, `4')dnl
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define(WORDSIZE, `32')dnl
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define(TOPBITS, eval(WORDSIZE - N*((WORDSIZE-1)/N)))dnl
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dnl
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define(dividend, `%o0')dnl
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define(divisor, `%o1')dnl
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define(Q, `%o2')dnl
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define(R, `%o3')dnl
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define(ITER, `%o4')dnl
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define(V, `%o5')dnl
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dnl
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dnl m4 reminder: ifelse(a,b,c,d) => if a is b, then c, else d
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define(T, `%g1')dnl
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define(SC, `%g7')dnl
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ifelse(S, `true', `define(SIGN, `%g6')')dnl
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dnl
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dnl This is the recursive definition for developing quotient digits.
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dnl
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dnl Parameters:
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dnl $1 the current depth, 1 <= $1 <= N
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dnl $2 the current accumulation of quotient bits
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dnl N max depth
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dnl
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dnl We add a new bit to $2 and either recurse or insert the bits in
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dnl the quotient. R, Q, and V are inputs and outputs as defined above;
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dnl the condition codes are expected to reflect the input R, and are
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dnl modified to reflect the output R.
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dnl
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define(DEVELOP_QUOTIENT_BITS,
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` ! depth $1, accumulated bits $2
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bl L.$1.eval(2**N+$2)
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srl V,1,V
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! remainder is positive
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subcc R,V,R
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ifelse($1, N,
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` b 9f
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add Q, ($2*2+1), Q
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', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2+1)')')
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L.$1.eval(2**N+$2):
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! remainder is negative
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addcc R,V,R
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ifelse($1, N,
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` b 9f
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add Q, ($2*2-1), Q
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', ` DEVELOP_QUOTIENT_BITS(incr($1), `eval(2*$2-1)')')
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ifelse($1, 1, `9:')')dnl
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#include "DEFS.h"
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#ifdef __svr4__
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#include <sys/trap.h>
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#else
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#include <machine/trap.h>
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#endif
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FUNC(NAME)
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ifelse(S, `true',
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` ! compute sign of result; if neither is negative, no problem
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orcc divisor, dividend, %g0 ! either negative?
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bge 2f ! no, go do the divide
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ifelse(OP, `div',
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` xor divisor, dividend, SIGN ! compute sign in any case',
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` mov dividend, SIGN ! sign of remainder matches dividend')
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tst divisor
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bge 1f
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tst dividend
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! divisor is definitely negative; dividend might also be negative
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bge 2f ! if dividend not negative...
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sub %g0, divisor, divisor ! in any case, make divisor nonneg
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1: ! dividend is negative, divisor is nonnegative
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sub %g0, dividend, dividend ! make dividend nonnegative
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2:
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')
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! Ready to divide. Compute size of quotient; scale comparand.
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orcc divisor, %g0, V
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bne 1f
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mov dividend, R
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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
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retl
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clr %o0
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1:
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cmp R, V ! if divisor exceeds dividend, done
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blu Lgot_result ! (and algorithm fails otherwise)
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clr Q
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sethi %hi(1 << (WORDSIZE - TOPBITS - 1)), T
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cmp R, T
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blu Lnot_really_big
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clr ITER
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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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1:
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cmp V, T
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bgeu 3f
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mov 1, SC
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sll V, N, V
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b 1b
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add ITER, 1, ITER
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! Now compute SC.
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2: addcc V, V, V
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bcc Lnot_too_big
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add SC, 1, SC
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! We get here if the divisor overflowed while shifting.
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! This means that R has the high-order bit set.
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! Restore V and subtract from R.
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sll T, TOPBITS, T ! high order bit
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srl V, 1, V ! rest of V
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add V, T, V
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b Ldo_single_div
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sub SC, 1, SC
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Lnot_too_big:
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3: cmp V, R
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blu 2b
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nop
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be Ldo_single_div
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nop
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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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! V > R: went too far: back up 1 step
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! srl V, 1, V
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! dec SC
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! do single-bit divide steps
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!
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! We have to be careful here. We know that R >= V, 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 R >= 0. Because both R and V 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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Ldo_single_div:
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subcc SC, 1, SC
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bl Lend_regular_divide
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nop
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sub R, V, R
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mov 1, Q
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b Lend_single_divloop
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nop
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Lsingle_divloop:
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sll Q, 1, Q
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bl 1f
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srl V, 1, V
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! R >= 0
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sub R, V, R
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b 2f
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add Q, 1, Q
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1: ! R < 0
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add R, V, R
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sub Q, 1, Q
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2:
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Lend_single_divloop:
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subcc SC, 1, SC
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bge Lsingle_divloop
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tst R
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b,a Lend_regular_divide
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Lnot_really_big:
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1:
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sll V, N, V
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cmp V, R
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bleu 1b
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addcc ITER, 1, ITER
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be Lgot_result
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sub ITER, 1, ITER
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tst R ! set up for initial iteration
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Ldivloop:
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sll Q, N, Q
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DEVELOP_QUOTIENT_BITS(1, 0)
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Lend_regular_divide:
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subcc ITER, 1, ITER
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bge Ldivloop
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tst R
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bl,a Lgot_result
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! non-restoring fixup here (one instruction only!)
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ifelse(OP, `div',
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` sub Q, 1, Q
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', ` add R, divisor, R
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')
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Lgot_result:
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ifelse(S, `true',
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` ! check to see if answer should be < 0
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tst SIGN
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bl,a 1f
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ifelse(OP, `div', `sub %g0, Q, Q', `sub %g0, R, R')
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1:')
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retl
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ifelse(OP, `div', `mov Q, %o0', `mov R, %o0')
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