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Replace .urem and .rem routines.

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Replace .urem and .rem routines.  Fix a problem
with signed operands and uses leaf procedure calling
convention.

From-SVN: r20090
This commit is contained in:
Catherine Moore 1998-05-27 08:40:31 -04:00
parent c111bb67de
commit 9718d50bb5
1 changed files with 298 additions and 156 deletions
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454
gcc/config/sparc/lb1spc.asm
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@ -282,205 +282,347 @@ got_result:
#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:
save %sp, -64, %sp
b divide
mov 0, %l2 ! result always positive
mov 0, %g3 ! result always positive
.align 4
.global .rem
.proc 4
.rem:
save %sp, -64, %sp
orcc %i1, %i0, %g0 ! is either operand negative
bge divide ! if not, skip this junk
mov %i0, %l2 ! record sign of result in sign of %i2
tst %i1
bge 2f
tst %i0
! %i1 < 0
bge divide
neg %i1
2: ! %i0 < 0
neg %i0
! FALL THROUGH
! 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:
! Compute size of quotient, scale comparand.
orcc %i1, %g0, %l1 ! movcc %i1, %l1
te 2 ! if %i1 = 0
mov %i0, %i3
mov 0, %i2
sethi %hi(1<<(32-4-1)), %l3
cmp %i3, %l3
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
mov 0, %l0
!
! Here, the %i0 is >= 2^(31-3) or so. We must be careful here,
! as our usual 3-at-a-shot divide step will cause overflow and havoc.
! The total number of bits in the result here is 3*%l0+%l4, where
! %l4 <= 3.
! Compute %l0 in an unorthodox manner: know we need to Shift %l1 into
! the top decade: so do not even bother to compare to %i3.
1: cmp %l1, %l3
bgeu 3f
mov 1, %l4
sll %l1, 3, %l1
b 1b
inc %l0
!
! Now compute %l4
!
2: addcc %l1, %l1, %l1
bcc not_too_big
add %l4, 1, %l4
!
! We are here if the %i1 overflowed when Shifting.
! This means that %i3 has the high-order bit set.
! Restore %l1 and subtract from %i3.
sll %l3, 4, %l3
srl %l1, 1, %l1
add %l1, %l3, %l1
b do_single_div
dec %l4
not_too_big:
3: cmp %l1, %i3
blu 2b
nop
be do_single_div
nop
! %l1 > %i3: went too far: back up 1 step
! srl %l1, 1, %l1
! dec %l4
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 %i3 >= %l1, so we can do the
! 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 %i3 >= 0. Because both %i3 and %l1 may have the
! high-order bit set in the first step, just falling into the regular
! 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 %l4
bl end_regular_divide
nop
sub %i3, %l1, %i3
mov 1, %i2
b end_single_divloop
nop
single_divloop:
sll %i2, 1, %i2
bl 1f
srl %l1, 1, %l1
! %i3 >= 0
sub %i3, %l1, %i3
b 2f
inc %i2
1: ! %i3 < 0
add %i3, %l1, %i3
dec %i2
end_single_divloop:
2: deccc %l4
bge single_divloop
tst %i3
b end_regular_divide
nop
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 %l1, 3, %l1
cmp %l1, %i3
1:
sll %o5, 4, %o5
cmp %o5, %o3
bleu 1b
inccc %l0
addcc %o4, 1, %o4
be got_result
dec %l0
do_regular_divide:
! Do the main division iteration
tst %i3
! Fall through into divide loop
sub %o4, 1, %o4
tst %o3 ! set up for initial iteration
divloop:
sll %i2, 3, %i2
! depth 1, accumulated bits 0
bl L.1.8
srl %l1,1,%l1
sll %o2, 4, %o2
! depth 1, accumulated bits 0
bl L1.16
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
subcc %o3,%o5,%o3
! depth 2, accumulated bits 1
bl L.2.9
srl %l1,1,%l1
bl L2.17
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
subcc %o3,%o5,%o3
! depth 3, accumulated bits 3
bl L.3.11
srl %l1,1,%l1
bl L3.19
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
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 %i2, (3*2+1), %i2
L.3.11: ! remainder is negative
addcc %i3,%l1,%i3
add %o2, (7*2+1), %o2
L4.23:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %i2, (3*2-1), %i2
L.2.9: ! remainder is negative
addcc %i3,%l1,%i3
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 L.3.9
srl %l1,1,%l1
bl L3.17
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
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 %i2, (1*2+1), %i2
L.3.9: ! remainder is negative
addcc %i3,%l1,%i3
add %o2, (3*2+1), %o2
L4.19:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %i2, (1*2-1), %i2
L.1.8: ! remainder is negative
addcc %i3,%l1,%i3
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 L.2.7
srl %l1,1,%l1
bl L2.15
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
subcc %o3,%o5,%o3
! depth 3, accumulated bits -1
bl L.3.7
srl %l1,1,%l1
bl L3.15
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
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 %i2, (-1*2+1), %i2
L.3.7: ! remainder is negative
addcc %i3,%l1,%i3
add %o2, (-1*2+1), %o2
L4.15:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %i2, (-1*2-1), %i2
L.2.7: ! remainder is negative
addcc %i3,%l1,%i3
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 L.3.5
srl %l1,1,%l1
bl L3.13
srl %o5,1,%o5
! remainder is positive
subcc %i3,%l1,%i3
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 %i2, (-3*2+1), %i2
L.3.5: ! remainder is negative
addcc %i3,%l1,%i3
add %o2, (-5*2+1), %o2
L4.11:
! remainder is negative
addcc %o3,%o5,%o3
b 9f
add %i2, (-3*2-1), %i2
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:
9: deccc %l0
subcc %o4, 1, %o4
bge divloop
tst %i3
bge got_result
nop
! non-restoring fixup here
add %i3, %i1, %i3
tst %o3
bl,a got_result
! non-restoring fixup here (one instruction only!)
add %o3, %o1, %o3
got_result:
tst %l2
bge 1f
restore
! answer < 0
retl ! leaf-routine return
neg %o3, %o0 ! remainder <- -%i3
1: retl ! leaf-routine return
mov %o3, %o0 ! remainder <- %i3
! check to see if answer should be < 0
tst %g3
bl,a 1f
sub %g0, %o3, %o3
1:
retl
mov %o3, %o0
#endif