/* This is an assembly language implementation of mulsi3, divsi3, and modsi3
for the sparc processor.
These routines are derived from the SPARC Architecture Manual, version 8,
slightly edited to match the desired calling convention, and also to
optimize them for our purposes. */
/* An executable stack is *not* required for these functions. */
#if defined(__ELF__) && defined(__linux__)
.section .note.GNU-stack,"",%progbits
.previous
#endif
#ifdef L_mulsi3
.text
.align 4
.global .umul
.proc 4
.umul:
or %o0, %o1, %o4 ! logical or of multiplier and multiplicand
mov %o0, %y ! multiplier to Y register
andncc %o4, 0xfff, %o5 ! mask out lower 12 bits
be mul_shortway ! can do it the short way
andcc %g0, %g0, %o4 ! zero the partial product and clear NV cc
!
! long multiply
!
mulscc %o4, %o1, %o4 ! first iteration of 33
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4 ! 32nd iteration
mulscc %o4, %g0, %o4 ! last iteration only shifts
! the upper 32 bits of product are wrong, but we do not care
retl
rd %y, %o0
!
! short multiply
!
mul_shortway:
mulscc %o4, %o1, %o4 ! first iteration of 13
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4
mulscc %o4, %o1, %o4 ! 12th iteration
mulscc %o4, %g0, %o4 ! last iteration only shifts
rd %y, %o5
sll %o4, 12, %o4 ! left shift partial product by 12 bits
srl %o5, 20, %o5 ! right shift partial product by 20 bits
retl
or %o5, %o4, %o0 ! merge for true product
#endif
#ifdef L_divsi3
/*
* 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.
*/
.global .udiv
.align 4
.proc 4
.text
.udiv:
b ready_to_divide
mov 0, %g3 ! result is always positive
.global .div
.align 4
.proc 4
.text
.div:
! compute sign of result; if neither is negative, no problem
orcc %o1, %o0, %g0 ! either negative?
bge ready_to_divide ! no, go do the divide
xor %o1, %o0, %g3 ! compute sign in any case
tst %o1
bge 1f
tst %o0
! %o1 is definitely negative; %o0 might also be negative
bge ready_to_divide ! 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
ready_to_divide:
! 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 0x2 ! ST_DIV0
retl
clr %o0
1:
cmp %o3, %o5 ! if %o1 exceeds %o0, done
blu got_result ! (and algorithm fails otherwise)
clr %o2
sethi %hi(1 << (32 - 4 - 1)), %g1
cmp %o3, %g1
blu not_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, %g2
sll %o5, 4, %o5
b 1b
add %o4, 1, %o4
! Now compute %g2.
2: addcc %o5, %o5, %o5
bcc not_too_big
add %g2, 1, %g2
! 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 do_single_div
sub %g2, 1, %g2
not_too_big:
3: cmp %o5, %o3
blu 2b
nop
be do_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 %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:
subcc %g2, 1, %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
add %o2, 1, %o2
1: ! %o3 < 0
add %o3, %o5, %o3
sub %o2, 1, %o2
2:
end_single_divloop:
subcc %g2, 1, %g2
bge single_divloop
tst %o3
b,a end_regular_divide
not_really_big:
1:
sll %o5, 4, %o5
cmp %o5, %o3
bleu 1b
addcc %o4, 1, %o4
be got_result
sub %o4, 1, %o4
tst %o3 ! set up for initial iteration
divloop:
sll %o2, 4, %o2
! depth 1, accumulated bits 0
bl L1.16
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 2, accumulated bits 1
bl L2.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits 3
bl L3.19
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 7
bl L4.23
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (7*2+1), %o2
L4.23:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (7*2-1), %o2
L3.19:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits 5
bl L4.21
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (5*2+1), %o2
L4.21:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (5*2-1), %o2
L2.17:
! remainder is negative
addcc %o3,%o5,%o3
! depth 3, accumulated bits 1
bl L3.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 3
bl L4.19
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (3*2+1), %o2
L4.19:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (3*2-1), %o2
L3.17:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits 1
bl L4.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (1*2+1), %o2
L4.17:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (1*2-1), %o2
L1.16:
! remainder is negative
addcc %o3,%o5,%o3
! depth 2, accumulated bits -1
bl L2.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits -1
bl L3.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -1
bl L4.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-1*2+1), %o2
L4.15:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-1*2-1), %o2
L3.15:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits -3
bl L4.13
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-3*2+1), %o2
L4.13:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-3*2-1), %o2
L2.15:
! remainder is negative
addcc %o3,%o5,%o3
! depth 3, accumulated bits -3
bl L3.13
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -5
bl L4.11
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-5*2+1), %o2
L4.11:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-5*2-1), %o2
L3.13:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits -7
bl L4.9
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-7*2+1), %o2
L4.9:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-7*2-1), %o2
9:
end_regular_divide:
subcc %o4, 1, %o4
bge divloop
tst %o3
bl,a got_result
! non-restoring fixup here (one instruction only!)
sub %o2, 1, %o2
got_result:
! check to see if answer should be < 0
tst %g3
bl,a 1f
sub %g0, %o2, %o2
1:
retl
mov %o2, %o0
#endif
#ifdef L_modsi3
/* This implementation was taken from glibc:
*
* Input: dividend and divisor in %o0 and %o1 respectively.
*
* 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.
*/
.text
.align 4
.global .urem
.proc 4
.urem:
b divide
mov 0, %g3 ! result always positive
.align 4
.global .rem
.proc 4
.rem:
! compute sign of result; if neither is negative, no problem
orcc %o1, %o0, %g0 ! either negative?
bge 2f ! no, go do the divide
mov %o0, %g3 ! sign of remainder matches %o0
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.
divide:
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 0x2 !ST_DIV0
retl
clr %o0
1:
cmp %o3, %o5 ! if %o1 exceeds %o0, done
blu got_result ! (and algorithm fails otherwise)
clr %o2
sethi %hi(1 << (32 - 4 - 1)), %g1
cmp %o3, %g1
blu not_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, %g2
sll %o5, 4, %o5
b 1b
add %o4, 1, %o4
! Now compute %g2.
2: addcc %o5, %o5, %o5
bcc not_too_big
add %g2, 1, %g2
! 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 do_single_div
sub %g2, 1, %g2
not_too_big:
3: cmp %o5, %o3
blu 2b
nop
be do_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 %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:
subcc %g2, 1, %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
add %o2, 1, %o2
1: ! %o3 < 0
add %o3, %o5, %o3
sub %o2, 1, %o2
2:
end_single_divloop:
subcc %g2, 1, %g2
bge single_divloop
tst %o3
b,a end_regular_divide
not_really_big:
1:
sll %o5, 4, %o5
cmp %o5, %o3
bleu 1b
addcc %o4, 1, %o4
be got_result
sub %o4, 1, %o4
tst %o3 ! set up for initial iteration
divloop:
sll %o2, 4, %o2
! depth 1, accumulated bits 0
bl L1.16
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 2, accumulated bits 1
bl L2.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits 3
bl L3.19
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 7
bl L4.23
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (7*2+1), %o2
L4.23:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (7*2-1), %o2
L3.19:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits 5
bl L4.21
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (5*2+1), %o2
L4.21:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (5*2-1), %o2
L2.17:
! remainder is negative
addcc %o3,%o5,%o3
! depth 3, accumulated bits 1
bl L3.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits 3
bl L4.19
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (3*2+1), %o2
L4.19:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (3*2-1), %o2
L3.17:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits 1
bl L4.17
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (1*2+1), %o2
L4.17:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (1*2-1), %o2
L1.16:
! remainder is negative
addcc %o3,%o5,%o3
! depth 2, accumulated bits -1
bl L2.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 3, accumulated bits -1
bl L3.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -1
bl L4.15
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-1*2+1), %o2
L4.15:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-1*2-1), %o2
L3.15:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits -3
bl L4.13
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-3*2+1), %o2
L4.13:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-3*2-1), %o2
L2.15:
! remainder is negative
addcc %o3,%o5,%o3
! depth 3, accumulated bits -3
bl L3.13
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
! depth 4, accumulated bits -5
bl L4.11
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-5*2+1), %o2
L4.11:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-5*2-1), %o2
L3.13:
! remainder is negative
addcc %o3,%o5,%o3
! depth 4, accumulated bits -7
bl L4.9
srl %o5,1,%o5
! remainder is positive
subcc %o3,%o5,%o3
b 9f
add %o2, (-7*2+1), %o2
L4.9:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %o2, (-7*2-1), %o2
9:
end_regular_divide:
subcc %o4, 1, %o4
bge divloop
tst %o3
bl,a got_result
! non-restoring fixup here (one instruction only!)
add %o3, %o1, %o3
got_result:
! check to see if answer should be < 0
tst %g3
bl,a 1f
sub %g0, %o3, %o3
1:
retl
mov %o3, %o0
#endif