Add SPARC64 version of div/mod written in assembly.

This version is similar to the code shipped with libgcc. It is based on
the code from the SPARC64 architecture manual, provided without any
restrictions.

Tested by:	flo@
This commit is contained in:
Ed Schouten 2012-01-12 17:55:22 +00:00
parent ce9f43b467
commit 848933e870
6 changed files with 924 additions and 0 deletions

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/*
* 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: define(N, 4) define(TWOSUPN, 16)
* WORDSIZE -- how many bits altogether we're talking about:
* obviously: define(WORDSIZE, 32)
* A derived constant:
* TOPBITS -- how many bits are in the top "decade" of a number:
* define(TOPBITS, eval( WORDSIZE - N*((WORDSIZE-1)/N) ) )
* 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)/N )
* Note that this is log_base_(2ˆN) of the quotient.
* V -- the current comparand -- initially divisor*2ˆ(ITER*N-1)
* Cost:
* current estimate for non-large dividend is
* CEIL( lg2(quotient) / N ) x ( 10 + 7N/2 ) + C
* a large dividend is one greater than 2ˆ(31-TOPBITS) 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.
*/
define(dividend, `%o0')
define(divisor,`%o1')
define(Q, `%o2')
define(R, `%o3')
define(ITER, `%o4')
define(V, `%o5')
define(SIGN, `%g3')
define(T, `%g1')
define(SC,`%g2')
/*
* This is the recursive definition of how we develop quotient digits.
* It takes three important parameters:
* $1 -- the current depth, 1<=$1<=N
* $2 -- the current accumulation of quotient bits
* N -- max depth
* We add a new bit to $2 and either recurse or insert the bits in the quotient.
* Dynamic input:
* R -- current remainder
* Q -- current quotient
* V -- current comparand
* cc -- set on current value of R
* Dynamic output:
* R', Q', V', cc'
*/
#include "../assembly.h"
.text
.align 4
define(DEVELOP_QUOTIENT_BITS,
` !depth $1, accumulated bits $2
bl L.$1.eval(TWOSUPN+$2)
srl V,1,V
! remainder is nonnegative
subcc R,V,R
ifelse( $1, N,
` b 9f
add Q, ($2*2+1), Q
',` DEVELOP_QUOTIENT_BITS( incr($1), `eval(2*$2+1)')
')
L.$1.eval(TWOSUPN+$2):
! remainder is negative
addcc R,V,R
ifelse( $1, N,
` b 9f
add Q, ($2*2-1), Q
',` DEVELOP_QUOTIENT_BITS( incr($1), `eval(2*$2-1)')
')
ifelse( $1, 1, `9:')
')
ifelse( ANSWER, `quotient', `
DEFINE_COMPILERRT_FUNCTION(__udivsi3)
save %sp,-64,%sp ! do this for debugging
b divide
mov 0,SIGN ! result always nonnegative
DEFINE_COMPILERRT_FUNCTION(__divsi3)
save %sp,-64,%sp ! do this for debugging
orcc divisor,dividend,%g0 ! are either dividend or divisor negative
bge divide ! if not, skip this junk
xor divisor,dividend,SIGN ! record sign of result in sign of SIGN
tst divisor
bge 2f
tst dividend
! divisor < 0
bge divide
neg divisor
2:
! dividend < 0
neg dividend
! FALL THROUGH
',`
DEFINE_COMPILERRT_FUNCTION(__umodsi3)
save %sp,-64,%sp ! do this for debugging
b divide
mov 0,SIGN ! result always nonnegative
DEFINE_COMPILERRT_FUNCTION(__modsi3)
save %sp,-64,%sp ! do this for debugging
orcc divisor,dividend,%g0 ! are either dividend or divisor negative
bge divide ! if not, skip this junk
mov dividend,SIGN ! record sign of result in sign of SIGN
tst divisor
bge 2f
tst dividend
! divisor < 0
bge divide
neg divisor
2:
! dividend < 0
neg dividend
! FALL THROUGH
')
divide:
! Compute size of quotient, scale comparand.
orcc divisor,%g0,V ! movcc divisor,V
te 2 ! if divisor = 0
mov dividend,R
mov 0,Q
sethi %hi(1<<(WORDSIZE-TOPBITS-1)),T
cmp R,T
blu not_really_big
mov 0,ITER
!
! Here, the dividend is >= 2ˆ(31-N) or so. We must be careful here,
! as our usual N-at-a-shot divide step will cause overflow and havoc.
! The total 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 don't even bother to compare to R.
1:
cmp V,T
bgeu 3f
mov 1,SC
sll V,N,V
b 1b
inc ITER
! Now compute SC
2: addcc V,V,V
bcc not_too_big
add SC,1,SC
! We're here if the divisor overflowed when Shifting.
! This means that R has the high-order bit set.
! Restore V and subtract from R.
sll T,TOPBITS,T ! high order bit
srl V,1,V ! rest of V
add V,T,V
b do_single_div
dec SC
not_too_big:
3: cmp V,R
blu 2b
nop
be do_single_div
nop
! V > R: went too far: back up 1 step
! srl V,1,V
! dec SC
! do single-bit divide steps
!
! We have to be careful here. We know that R >= V, so we can do the
! first divide step without thinking. BUT, the others are conditional,
! and are only done if R >= 0. Because both R and V 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 SC
bl end_regular_divide
nop
sub R,V,R
mov 1,Q
b end_single_divloop
nop
single_divloop:
sll Q,1,Q
bl 1f
srl V,1,V
! R >= 0
sub R,V,R
b 2f
inc Q
1: ! R < 0
add R,V,R
dec Q
2:
end_single_divloop:
deccc SC
bge single_divloop
tst R
b end_regular_divide
nop
not_really_big:
1:
sll V,N,V
cmp V,R
bleu 1b
inccc ITER
be got_result
dec ITER
do_regular_divide:
! Do the main division iteration
tst R
! Fall through into divide loop
divloop:
sll Q,N,Q
DEVELOP_QUOTIENT_BITS( 1, 0 )
end_regular_divide:
deccc ITER
bge divloop
tst R
bge got_result
nop
! non-restoring fixup here
ifelse( ANSWER, `quotient',
` dec Q
',` add R,divisor,R
')
got_result:
tst SIGN
bge 1f
restore
! answer < 0
retl ! leaf-routine return
ifelse( ANSWER, `quotient',
` neg %o2,%o0 ! quotient <- -Q
',` neg %o3,%o0 ! remainder <- -R
')
1:
retl ! leaf-routine return
ifelse( ANSWER, `quotient',
` mov %o2,%o0 ! quotient <- Q
',` mov %o3,%o0 ! remainder <- R
')

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/*
* 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 4
DEFINE_COMPILERRT_FUNCTION(__udivsi3)
save %sp,-64,%sp ! do this for debugging
b divide
mov 0,%g3 ! result always nonnegative
DEFINE_COMPILERRT_FUNCTION(__divsi3)
save %sp,-64,%sp ! do this for debugging
orcc %o1,%o0,%g0 ! are either %o0 or %o1 negative
bge divide ! if not, skip this junk
xor %o1,%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 end_single_divloop
nop
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 end_regular_divide
nop
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
bge got_result
nop
! non-restoring fixup here
dec %o2
got_result:
tst %g3
bge 1f
restore
! answer < 0
retl ! leaf-routine return
neg %o2,%o0 ! quotient <- -%o2
1:
retl ! leaf-routine return
mov %o2,%o0 ! quotient <- %o2

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#!/bin/sh
m4 divmod.m4 | sed -e 's/[[:space:]]*$//' | grep -v '^$' > modsi3.S
m4 -DANSWER=quotient divmod.m4 | sed -e 's/[[:space:]]*$//' | grep -v '^$' > divsi3.S
echo '! This file intentionally left blank' > umodsi3.S
echo '! This file intentionally left blank' > udivsi3.S

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/*
* 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 4
DEFINE_COMPILERRT_FUNCTION(__umodsi3)
save %sp,-64,%sp ! do this for debugging
b divide
mov 0,%g3 ! result always nonnegative
DEFINE_COMPILERRT_FUNCTION(__modsi3)
save %sp,-64,%sp ! do this for debugging
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 end_single_divloop
nop
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 end_regular_divide
nop
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
bge got_result
nop
! non-restoring fixup here
add %o3,%o1,%o3
got_result:
tst %g3
bge 1f
restore
! answer < 0
retl ! leaf-routine return
neg %o3,%o0 ! remainder <- -%o3
1:
retl ! leaf-routine return
mov %o3,%o0 ! remainder <- %o3

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