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dnl Alpha mpn_modexact_1c_odd -- mpn exact remainder
dnl Copyright 2003, 2004 Free Software Foundation, Inc.
dnl
dnl This file is part of the GNU MP Library.
dnl
dnl The GNU MP Library is free software; you can redistribute it and/or
dnl modify it under the terms of the GNU Lesser General Public License as
dnl published by the Free Software Foundation; either version 3 of the
dnl License, or (at your option) any later version.
dnl
dnl The GNU MP Library is distributed in the hope that it will be useful,
dnl but WITHOUT ANY WARRANTY; without even the implied warranty of
dnl MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
dnl Lesser General Public License for more details.
dnl
dnl You should have received a copy of the GNU Lesser General Public License
dnl along with the GNU MP Library. If not, see http://www.gnu.org/licenses/.
include(`../config.m4')
C cycles/limb
C EV4: 47
C EV5: 30
C EV6: 15
C mp_limb_t mpn_modexact_1c_odd (mp_srcptr src, mp_size_t size, mp_limb_t d,
C mp_limb_t c)
C
C This code follows the "alternate" code in mpn/generic/mode1o.c,
C eliminating cbit+climb from the dependent chain. This leaves,
C
C ev4 ev5 ev6
C 1 3 1 subq y = x - h
C 23 13 7 mulq q = y * inverse
C 23 14 7 umulh h = high (q * d)
C -- -- --
C 47 30 15
C
C In each case, the load latency, loop control, and extra carry bit handling
C hide under the multiply latencies. Those latencies are long enough that
C we don't need to worry about alignment or pairing to squeeze out
C performance.
C
C For the first limb, some of the loop code is broken out and scheduled back
C since it can be done earlier.
C
C - The first ldq src[0] is near the start of the routine, for maximum
C time from memory.
C
C - The subq y=x-climb can be done without waiting for the inverse.
C
C - The mulq y*inverse is replicated after the final subq for the inverse,
C instead of branching to the mulq in the main loop. On ev4 a branch
C there would cost cycles, but we can hide them under the mulq latency.
C
C For the last limb, high<divisor is tested and if that's true a subtract
C and addback is done, as per the main mpn/generic/mode1o.c code. This is a
C data-dependent branch, but we're waiting for umulh so any penalty should
C hide there. The multiplies saved would be worth the cost anyway.
C
C Enhancements:
C
C For size==1, a plain division (done bitwise say) might be faster than
C calculating an inverse, the latter taking about 130 cycles on ev4 or 70 on
C ev5. A call to gcc __remqu might be a possibility.
ASM_START()
PROLOGUE(mpn_modexact_1c_odd,gp)
C r16 src
C r17 size
C r18 d
C r19 c
LEA(r0, binvert_limb_table)
srl r18, 1, r20 C d >> 1
and r20, 127, r20 C idx = d>>1 & 0x7F
addq r0, r20, r21 C table + idx
ifelse(bwx_available_p,1,
` ldbu r20, 0(r21) C table[idx], inverse 8 bits
',`
ldq_u r20, 0(r21) C table[idx] qword
extbl r20, r21, r20 C table[idx], inverse 8 bits
')
mull r20, r20, r7 C i*i
addq r20, r20, r20 C 2*i
ldq r2, 0(r16) C x = s = src[0]
lda r17, -1(r17) C size--
clr r0 C initial cbit=0
mull r7, r18, r7 C i*i*d
subq r20, r7, r20 C 2*i-i*i*d, inverse 16 bits
mull r20, r20, r7 C i*i
addq r20, r20, r20 C 2*i
mull r7, r18, r7 C i*i*d
subq r20, r7, r20 C 2*i-i*i*d, inverse 32 bits
mulq r20, r20, r7 C i*i
addq r20, r20, r20 C 2*i
mulq r7, r18, r7 C i*i*d
subq r2, r19, r3 C y = x - climb
subq r20, r7, r20 C inv = 2*i-i*i*d, inverse 64 bits
ASSERT(r7, C should have d*inv==1 mod 2^64
` mulq r18, r20, r7
cmpeq r7, 1, r7')
mulq r3, r20, r4 C first q = y * inv
beq r17, L(one) C if size==1
br L(entry)
L(top):
C r0 cbit
C r16 src, incrementing
C r17 size, decrementing
C r18 d
C r19 climb
C r20 inv
ldq r1, 0(r16) C s = src[i]
subq r1, r0, r2 C x = s - cbit
cmpult r1, r0, r0 C new cbit = s < cbit
subq r2, r19, r3 C y = x - climb
mulq r3, r20, r4 C q = y * inv
L(entry):
cmpult r2, r19, r5 C cbit2 = x < climb
addq r5, r0, r0 C cbit += cbit2
lda r16, 8(r16) C src++
lda r17, -1(r17) C size--
umulh r4, r18, r19 C climb = q * d
bne r17, L(top) C while 2 or more limbs left
C r0 cbit
C r18 d
C r19 climb
C r20 inv
ldq r1, 0(r16) C s = src[size-1] high limb
cmpult r1, r18, r2 C test high<divisor
bne r2, L(skip) C skip if so
C can't skip a division, repeat loop code
subq r1, r0, r2 C x = s - cbit
cmpult r1, r0, r0 C new cbit = s < cbit
subq r2, r19, r3 C y = x - climb
mulq r3, r20, r4 C q = y * inv
L(one):
cmpult r2, r19, r5 C cbit2 = x < climb
addq r5, r0, r0 C cbit += cbit2
umulh r4, r18, r19 C climb = q * d
addq r19, r0, r0 C return climb + cbit
ret r31, (r26), 1
ALIGN(8)
L(skip):
C with high<divisor, the final step can be just (cbit+climb)-s and
C an addback of d if that underflows
addq r19, r0, r19 C c = climb + cbit
subq r19, r1, r2 C c - s
cmpult r19, r1, r3 C c < s
addq r2, r18, r0 C return c-s + divisor
cmoveq r3, r2, r0 C return c-s if no underflow
ret r31, (r26), 1
EPILOGUE()
ASM_END()