266 lines
6.2 KiB
ArmAsm
266 lines
6.2 KiB
ArmAsm
/* Copyright (C) 2004 Free Software Foundation, Inc.
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This file is part of the GNU C Library.
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The GNU C Library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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The GNU C Library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with the GNU C Library; if not, write to the Free
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Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
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02111-1307 USA. */
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#include "div_libc.h"
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/* 64-bit signed long divide. These are not normal C functions. Argument
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registers are t10 and t11, the result goes in t12. Only t12 and AT may
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be clobbered.
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Theory of operation here is that we can use the FPU divider for virtually
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all operands that we see: all dividend values between -2**53 and 2**53-1
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can be computed directly. Note that divisor values need not be checked
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against that range because the rounded fp value will be close enough such
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that the quotient is < 1, which will properly be truncated to zero when we
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convert back to integer.
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When the dividend is outside the range for which we can compute exact
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results, we use the fp quotent as an estimate from which we begin refining
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an exact integral value. This reduces the number of iterations in the
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shift-and-subtract loop significantly. */
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.text
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.align 4
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.globl __divq
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.type __divq, @function
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.usepv __divq, no
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cfi_startproc
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cfi_return_column (RA)
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__divq:
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lda sp, -FRAME(sp)
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cfi_def_cfa_offset (FRAME)
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CALL_MCOUNT
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/* Get the fp divide insn issued as quickly as possible. After
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that's done, we have at least 22 cycles until its results are
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ready -- all the time in the world to figure out how we're
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going to use the results. */
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stq X, 16(sp)
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stq Y, 24(sp)
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beq Y, DIVBYZERO
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stt $f0, 0(sp)
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stt $f1, 8(sp)
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cfi_rel_offset ($f0, 0)
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cfi_rel_offset ($f1, 8)
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ldt $f0, 16(sp)
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ldt $f1, 24(sp)
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cvtqt $f0, $f0
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cvtqt $f1, $f1
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divt/c $f0, $f1, $f0
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/* Check to see if X fit in the double as an exact value. */
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sll X, (64-53), AT
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ldt $f1, 8(sp)
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sra AT, (64-53), AT
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cmpeq X, AT, AT
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beq AT, $x_big
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/* If we get here, we're expecting exact results from the division.
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Do nothing else besides convert and clean up. */
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cvttq/c $f0, $f0
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stt $f0, 16(sp)
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ldq RV, 16(sp)
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ldt $f0, 0(sp)
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cfi_restore ($f1)
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cfi_remember_state
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cfi_restore ($f0)
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cfi_def_cfa_offset (0)
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lda sp, FRAME(sp)
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ret $31, (RA), 1
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.align 4
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cfi_restore_state
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$x_big:
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/* If we get here, X is large enough that we don't expect exact
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results, and neither X nor Y got mis-translated for the fp
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division. Our task is to take the fp result, figure out how
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far it's off from the correct result and compute a fixup. */
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stq t0, 16(sp)
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stq t1, 24(sp)
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stq t2, 32(sp)
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stq t5, 40(sp)
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cfi_rel_offset (t0, 16)
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cfi_rel_offset (t1, 24)
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cfi_rel_offset (t2, 32)
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cfi_rel_offset (t5, 40)
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#define Q RV /* quotient */
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#define R t0 /* remainder */
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#define SY t1 /* scaled Y */
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#define S t2 /* scalar */
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#define QY t3 /* Q*Y */
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/* The fixup code below can only handle unsigned values. */
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or X, Y, AT
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mov $31, t5
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blt AT, $fix_sign_in
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$fix_sign_in_ret1:
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cvttq/c $f0, $f0
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stt $f0, 8(sp)
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ldq Q, 8(sp)
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$fix_sign_in_ret2:
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mulq Q, Y, QY
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stq t4, 8(sp)
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ldt $f0, 0(sp)
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unop
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cfi_rel_offset (t4, 8)
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cfi_restore ($f0)
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stq t3, 0(sp)
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unop
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cfi_rel_offset (t3, 0)
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subq QY, X, R
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mov Y, SY
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mov 1, S
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bgt R, $q_high
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$q_high_ret:
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subq X, QY, R
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mov Y, SY
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mov 1, S
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bgt R, $q_low
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$q_low_ret:
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ldq t0, 16(sp)
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ldq t1, 24(sp)
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ldq t2, 32(sp)
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bne t5, $fix_sign_out
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$fix_sign_out_ret:
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ldq t3, 0(sp)
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ldq t4, 8(sp)
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ldq t5, 40(sp)
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lda sp, FRAME(sp)
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cfi_remember_state
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cfi_restore (t0)
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cfi_restore (t1)
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cfi_restore (t2)
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cfi_restore (t3)
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cfi_restore (t4)
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cfi_restore (t5)
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cfi_def_cfa_offset (0)
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ret $31, (RA), 1
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.align 4
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cfi_restore_state
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/* The quotient that we computed was too large. We need to reduce
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it by S such that Y*S >= R. Obviously the closer we get to the
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correct value the better, but overshooting high is ok, as we'll
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fix that up later. */
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0:
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addq SY, SY, SY
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addq S, S, S
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$q_high:
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cmpult SY, R, AT
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bne AT, 0b
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subq Q, S, Q
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unop
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subq QY, SY, QY
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br $q_high_ret
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.align 4
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/* The quotient that we computed was too small. Divide Y by the
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current remainder (R) and add that to the existing quotient (Q).
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The expectation, of course, is that R is much smaller than X. */
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/* Begin with a shift-up loop. Compute S such that Y*S >= R. We
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already have a copy of Y in SY and the value 1 in S. */
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0:
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addq SY, SY, SY
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addq S, S, S
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$q_low:
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cmpult SY, R, AT
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bne AT, 0b
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/* Shift-down and subtract loop. Each iteration compares our scaled
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Y (SY) with the remainder (R); if SY <= R then X is divisible by
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Y's scalar (S) so add it to the quotient (Q). */
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2: addq Q, S, t3
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srl S, 1, S
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cmpule SY, R, AT
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subq R, SY, t4
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cmovne AT, t3, Q
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cmovne AT, t4, R
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srl SY, 1, SY
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bne S, 2b
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br $q_low_ret
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.align 4
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$fix_sign_in:
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/* If we got here, then X|Y is negative. Need to adjust everything
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such that we're doing unsigned division in the fixup loop. */
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/* T5 records the changes we had to make:
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bit 0: set if result should be negative.
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bit 2: set if X was negated.
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bit 3: set if Y was negated.
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*/
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xor X, Y, AT
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cmplt AT, 0, t5
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cmplt X, 0, AT
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negq X, t0
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s4addq AT, t5, t5
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cmovne AT, t0, X
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cmplt Y, 0, AT
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negq Y, t0
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s8addq AT, t5, t5
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cmovne AT, t0, Y
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unop
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blbc t5, $fix_sign_in_ret1
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cvttq/c $f0, $f0
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stt $f0, 8(sp)
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ldq Q, 8(sp)
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unop
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negq Q, Q
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br $fix_sign_in_ret2
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.align 4
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$fix_sign_out:
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/* Now we get to undo what we did above. */
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/* ??? Is this really faster than just increasing the size of
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the stack frame and storing X and Y in memory? */
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and t5, 8, AT
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negq Y, t4
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cmovne AT, t4, Y
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and t5, 4, AT
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negq X, t4
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cmovne AT, t4, X
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negq RV, t4
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cmovlbs t5, t4, RV
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br $fix_sign_out_ret
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cfi_endproc
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.size __divq, .-__divq
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DO_DIVBYZERO
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