/* $Id: sdiv.S,v 1.6 1996/10/02 17:37:00 davem Exp $
* sdiv.S: This routine was taken from glibc-1.09 and is covered
* by the GNU Library General Public License Version 2.
*/
/* This file is generated from divrem.m4; DO NOT EDIT! */
/*
* Division and remainder, from Appendix E of the Sparc Version 8
* Architecture Manual, with fixes from Gordon Irlam.
*/
/*
* Input: dividend and divisor in %o0 and %o1 respectively.
*
* m4 parameters:
* .div name of function to generate
* div div=div => %o0 / %o1; div=rem => %o0 % %o1
* true true=true => signed; true=false => unsigned
*
* Algorithm parameters:
* N how many bits per iteration we try to get (4)
* WORDSIZE total number of bits (32)
*
* Derived constants:
* TOPBITS number of bits in the top decade of a number
*
* Important variables:
* Q the partial quotient under development (initially 0)
* R the remainder so far, initially the dividend
* ITER number of main division loop iterations required;
* equal to ceil(log2(quotient) / N). Note that this
* is the log base (2^N) of the quotient.
* V the current comparand, initially divisor*2^(ITER*N-1)
*
* Cost:
* Current estimate for non-large dividend is
* ceil(log2(quotient) / N) * (10 + 7N/2) + C
* A large dividend is one greater than 2^(31-TOPBITS) and takes a
* different path, as the upper bits of the quotient must be developed
* one bit at a time.
*/
.globl .div
.globl _Div
.div:
_Div: /* needed for export */
! compute sign of result; if neither is negative, no problem
orcc %o1, %o0, %g0 ! either negative?
bge 2f ! no, go do the divide
xor %o1, %o0, %g2 ! compute sign in any case
tst %o1
bge 1f
tst %o0
! %o1 is definitely negative; %o0 might also be negative
bge 2f ! if %o0 not negative...
sub %g0, %o1, %o1 ! in any case, make %o1 nonneg
1: ! %o0 is negative, %o1 is nonnegative
sub %g0, %o0, %o0 ! make %o0 nonnegative
2:
! Ready to divide. Compute size of quotient; scale comparand.
orcc %o1, %g0, %o5
bne 1f
mov %o0, %o3
! Divide by zero trap. If it returns, return 0 (about as
! wrong as possible, but that is what SunOS does...).
ta ST_DIV0
retl
clr %o0
1:
cmp %o3, %o5 ! if %o1 exceeds %o0, done
blu Lgot_result ! (and algorithm fails otherwise)
clr %o2
sethi %hi(1 << (32 - 4 - 1)), %g1
cmp %o3, %g1
blu Lnot_really_big
clr %o4
! Here the dividend is >= 2**(31-N) or so. We must be careful here,
! as our usual N-at-a-shot divide step will cause overflow and havoc.
! The 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 do not even bother to compare to R.
1:
cmp %o5, %g1
bgeu 3f
mov 1, %g7
sll %o5, 4, %o5
b 1b
add %o4, 1, %o4
! Now compute %g7.
2:
addcc %o5, %o5, %o5
bcc Lnot_too_big
add %g7, 1, %g7
! We get here if the %o1 overflowed while 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 Ldo_single_div
sub %g7, 1, %g7
Lnot_too_big:
3:
cmp %o5, %o3
blu 2b
nop
be Ldo_single_div
nop
/* NB: these are commented out in the V8-Sparc manual as well */
/* (I do not understand this) */
! %o5 > %o3: went too far: back up 1 step
! srl %o5, 1, %o5
! dec %g7
! 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...
Ldo_single_div:
subcc %g7, 1, %g7
bl Lend_regular_divide
nop
sub %o3, %o5, %o3
mov 1, %o2
b Lend_single_divloop
nop
Lsingle_divloop:
sll %o2, 1, %o2
bl 1f
srl %o5, 1, %o5
! %o3 >= 0
sub %o3, %o5, %o3
b 2f
add %o2, 1, %o2
1: ! %o3 < 0
add %o3, %o5, %o3
sub %o2, 1, %o2
2:
Lend_single_divloop:
subcc %g7, 1, %g7
bge Lsingle_divloop
tst %o3
b,a Lend_regular_divide
Lnot_really_big:
1:
sll %o5, 4, %o5
cmp %o5, %o3
bleu 1b
addcc %o4, 1, %o4
be Lgot_result
sub %o4, 1, %o4
tst %o3 ! set up for initial iteration
Ldivloop:
sll %o2, 4, %o2
! depth 1, accumulated bits 0
bl L.1.16
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 2, accumulated bits 1
bl L.2.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits 3
bl L.3.19
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 7
bl L.4.23
srl %o5,1,%o5
! remainder is positive
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 positive
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 positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 3
bl L.4.19
srl %o5,1,%o5
! remainder is positive
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 positive
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 positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits -1
bl L.3.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -1
bl L.4.15
srl %o5,1,%o5
! remainder is positive
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 positive
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 positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -5
bl L.4.11
srl %o5,1,%o5
! remainder is positive
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 positive
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:
Lend_regular_divide:
subcc %o4, 1, %o4
bge Ldivloop
tst %o3
bl,a Lgot_result
! non-restoring fixup here (one instruction only!)
sub %o2, 1, %o2
Lgot_result:
! check to see if answer should be < 0
tst %g2
bl,a 1f
sub %g0, %o2, %o2
1:
retl
mov %o2, %o0
.globl .div_patch
.div_patch:
sra %o0, 0x1f, %o2
wr %o2, 0x0, %y
nop
nop
nop
sdivcc %o0, %o1, %o0
bvs,a 1f
xnor %o0, %g0, %o0
1: retl
nop
