glibc/sysdeps/alpha/remq.S

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/* Copyright (C) 2004 Free Software Foundation, Inc.
This file is part of the GNU C Library.
1995-02-17 20:14:40 +00:00
The GNU C Library is free software; you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public
License as published by the Free Software Foundation; either
version 2.1 of the License, or (at your option) any later version.
The GNU C Library is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
Lesser General Public License for more details.
You should have received a copy of the GNU Lesser General Public
License along with the GNU C Library; if not, write to the Free
Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
02111-1307 USA. */
#include "div_libc.h"
/* 64-bit signed long remainder. These are not normal C functions. Argument
registers are t10 and t11, the result goes in t12. Only t12 and AT may
be clobbered.
Theory of operation here is that we can use the FPU divider for virtually
all operands that we see: all dividend values between -2**53 and 2**53-1
can be computed directly. Note that divisor values need not be checked
against that range because the rounded fp value will be close enough such
that the quotient is < 1, which will properly be truncated to zero when we
convert back to integer.
When the dividend is outside the range for which we can compute exact
results, we use the fp quotent as an estimate from which we begin refining
an exact integral value. This reduces the number of iterations in the
shift-and-subtract loop significantly. */
.text
.align 4
.globl __remq
.type __remq, @function
.usepv __remq, no
cfi_startproc
cfi_return_column (RA)
__remq:
lda sp, -FRAME(sp)
cfi_def_cfa_offset (FRAME)
CALL_MCOUNT
/* Get the fp divide insn issued as quickly as possible. After
that's done, we have at least 22 cycles until its results are
ready -- all the time in the world to figure out how we're
going to use the results. */
stt $f0, 0(sp)
stt $f1, 8(sp)
beq Y, DIVBYZERO
cfi_rel_offset ($f0, 0)
cfi_rel_offset ($f1, 8)
_ITOFT2 X, $f0, 16, Y, $f1, 24
cvtqt $f0, $f0
cvtqt $f1, $f1
divt/c $f0, $f1, $f0
/* Check to see if X fit in the double as an exact value. */
sll X, (64-53), AT
ldt $f1, 8(sp)
sra AT, (64-53), AT
cmpeq X, AT, AT
beq AT, $x_big
/* If we get here, we're expecting exact results from the division.
Do nothing else besides convert, compute remainder, clean up. */
cvttq/c $f0, $f0
_FTOIT $f0, AT, 16
mulq AT, Y, AT
ldt $f0, 0(sp)
cfi_restore ($f1)
cfi_remember_state
cfi_restore ($f0)
cfi_def_cfa_offset (0)
lda sp, FRAME(sp)
subq X, AT, RV
ret $31, (RA), 1
.align 4
cfi_restore_state
$x_big:
/* If we get here, X is large enough that we don't expect exact
results, and neither X nor Y got mis-translated for the fp
division. Our task is to take the fp result, figure out how
far it's off from the correct result and compute a fixup. */
stq t0, 16(sp)
stq t1, 24(sp)
stq t2, 32(sp)
stq t5, 40(sp)
cfi_rel_offset (t0, 16)
cfi_rel_offset (t1, 24)
cfi_rel_offset (t2, 32)
cfi_rel_offset (t5, 40)
#define Q t0 /* quotient */
#define R RV /* remainder */
#define SY t1 /* scaled Y */
#define S t2 /* scalar */
#define QY t3 /* Q*Y */
/* The fixup code below can only handle unsigned values. */
or X, Y, AT
mov $31, t5
blt AT, $fix_sign_in
$fix_sign_in_ret1:
cvttq/c $f0, $f0
_FTOIT $f0, Q, 8
.align 3
$fix_sign_in_ret2:
mulq Q, Y, QY
stq t4, 8(sp)
ldt $f0, 0(sp)
unop
cfi_rel_offset (t4, 8)
cfi_restore ($f0)
stq t3, 0(sp)
unop
cfi_rel_offset (t3, 0)
subq QY, X, R
mov Y, SY
mov 1, S
bgt R, $q_high
$q_high_ret:
subq X, QY, R
mov Y, SY
mov 1, S
bgt R, $q_low
$q_low_ret:
ldq t0, 16(sp)
ldq t1, 24(sp)
ldq t2, 32(sp)
bne t5, $fix_sign_out
$fix_sign_out_ret:
ldq t3, 0(sp)
ldq t4, 8(sp)
ldq t5, 40(sp)
lda sp, FRAME(sp)
cfi_remember_state
cfi_restore (t0)
cfi_restore (t1)
cfi_restore (t2)
cfi_restore (t3)
cfi_restore (t4)
cfi_restore (t5)
cfi_def_cfa_offset (0)
ret $31, (RA), 1
.align 4
cfi_restore_state
/* The quotient that we computed was too large. We need to reduce
it by S such that Y*S >= R. Obviously the closer we get to the
correct value the better, but overshooting high is ok, as we'll
fix that up later. */
0:
addq SY, SY, SY
addq S, S, S
$q_high:
cmpult SY, R, AT
bne AT, 0b
subq Q, S, Q
unop
subq QY, SY, QY
br $q_high_ret
.align 4
/* The quotient that we computed was too small. Divide Y by the
current remainder (R) and add that to the existing quotient (Q).
The expectation, of course, is that R is much smaller than X. */
/* Begin with a shift-up loop. Compute S such that Y*S >= R. We
already have a copy of Y in SY and the value 1 in S. */
0:
addq SY, SY, SY
addq S, S, S
$q_low:
cmpult SY, R, AT
bne AT, 0b
/* Shift-down and subtract loop. Each iteration compares our scaled
Y (SY) with the remainder (R); if SY <= R then X is divisible by
Y's scalar (S) so add it to the quotient (Q). */
2: addq Q, S, t3
srl S, 1, S
cmpule SY, R, AT
subq R, SY, t4
cmovne AT, t3, Q
cmovne AT, t4, R
srl SY, 1, SY
bne S, 2b
br $q_low_ret
.align 4
$fix_sign_in:
/* If we got here, then X|Y is negative. Need to adjust everything
such that we're doing unsigned division in the fixup loop. */
/* T5 records the changes we had to make:
bit 0: set if X was negated. Note that the sign of the
remainder follows the sign of the divisor.
bit 2: set if Y was negated.
*/
xor X, Y, t1
cmplt X, 0, t5
negq X, t0
cmovne t5, t0, X
cmplt Y, 0, AT
negq Y, t0
s4addq AT, t5, t5
cmovne AT, t0, Y
bge t1, $fix_sign_in_ret1
cvttq/c $f0, $f0
_FTOIT $f0, Q, 8
.align 3
negq Q, Q
br $fix_sign_in_ret2
.align 4
$fix_sign_out:
/* Now we get to undo what we did above. */
/* ??? Is this really faster than just increasing the size of
the stack frame and storing X and Y in memory? */
and t5, 4, AT
negq Y, t4
cmovne AT, t4, Y
negq X, t4
cmovlbs t5, t4, X
negq RV, t4
cmovlbs t5, t4, RV
br $fix_sign_out_ret
cfi_endproc
.size __remq, .-__remq
DO_DIVBYZERO