Commit 89a6fe72 authored by Niels Möller's avatar Niels Möller

arm: Moved memxor3 to new file, arm/memxor3.asm.

parent 93694133
2015-01-10 Niels Möller <nisse@lysator.liu.se>
* arm/memxor.asm (memxor3): Moved to new file.
* arm/memxor3.asm: New file.
2014-11-24 Niels Möller <nisse@lysator.liu.se> 2014-11-24 Niels Möller <nisse@lysator.liu.se>
* x86_64/memxor3.asm (memxor3): New file, code moved from old * x86_64/memxor3.asm (memxor3): New file, code moved from old
......
...@@ -227,274 +227,3 @@ PROLOGUE(nettle_memxor) ...@@ -227,274 +227,3 @@ PROLOGUE(nettle_memxor)
b .Lmemxor_bytes b .Lmemxor_bytes
EPILOGUE(nettle_memxor) EPILOGUE(nettle_memxor)
define(<DST>, <r0>)
define(<AP>, <r1>)
define(<BP>, <r2>)
define(<N>, <r3>)
undefine(<CNT>)
undefine(<TNC>)
C Temporaries r4-r7
define(<ACNT>, <r8>)
define(<ATNC>, <r10>)
define(<BCNT>, <r11>)
define(<BTNC>, <r12>)
C memxor3(void *dst, const void *a, const void *b, size_t n)
.align 2
PROLOGUE(nettle_memxor3)
cmp N, #0
beq .Lmemxor3_ret
push {r4,r5,r6,r7,r8,r10,r11}
cmp N, #7
add AP, N
add BP, N
add DST, N
bcs .Lmemxor3_large
C Simple byte loop
.Lmemxor3_bytes:
ldrb r4, [AP, #-1]!
ldrb r5, [BP, #-1]!
eor r4, r5
strb r4, [DST, #-1]!
subs N, #1
bne .Lmemxor3_bytes
.Lmemxor3_done:
pop {r4,r5,r6,r7,r8,r10,r11}
.Lmemxor3_ret:
bx lr
.Lmemxor3_align_loop:
ldrb r4, [AP, #-1]!
ldrb r5, [BP, #-1]!
eor r5, r4
strb r5, [DST, #-1]!
sub N, #1
.Lmemxor3_large:
tst DST, #3
bne .Lmemxor3_align_loop
C We have at least 4 bytes left to do here.
sub N, #4
ands ACNT, AP, #3
lsl ACNT, #3
beq .Lmemxor3_a_aligned
ands BCNT, BP, #3
lsl BCNT, #3
bne .Lmemxor3_uu
C Swap
mov r4, AP
mov AP, BP
mov BP, r4
.Lmemxor3_au:
C NOTE: We have the relevant shift count in ACNT, not BCNT
C AP is aligned, BP is not
C v original SRC
C +-------+------+
C |SRC-4 |SRC |
C +---+---+------+
C |DST-4 |
C +-------+
C
C With little-endian, we need to do
C DST[i-i] ^= (SRC[i-i] >> CNT) ^ (SRC[i] << TNC)
rsb ATNC, ACNT, #32
bic BP, #3
ldr r4, [BP]
tst N, #4
itet eq
moveq r5, r4
subne N, #4
beq .Lmemxor3_au_odd
.Lmemxor3_au_loop:
ldr r5, [BP, #-4]!
ldr r6, [AP, #-4]!
eor r6, r6, r4, lsl ATNC
eor r6, r6, r5, lsr ACNT
str r6, [DST, #-4]!
.Lmemxor3_au_odd:
ldr r4, [BP, #-4]!
ldr r6, [AP, #-4]!
eor r6, r6, r5, lsl ATNC
eor r6, r6, r4, lsr ACNT
str r6, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_au_loop
adds N, #8
beq .Lmemxor3_done
C Leftover bytes in r4, low end
ldr r5, [AP, #-4]
eor r4, r5, r4, lsl ATNC
.Lmemxor3_au_leftover:
C Store a byte at a time
ror r4, #24
strb r4, [DST, #-1]!
subs N, #1
beq .Lmemxor3_done
subs ACNT, #8
sub AP, #1
bne .Lmemxor3_au_leftover
b .Lmemxor3_bytes
.Lmemxor3_a_aligned:
ands ACNT, BP, #3
lsl ACNT, #3
bne .Lmemxor3_au ;
C a, b and dst all have the same alignment.
subs N, #8
bcc .Lmemxor3_aligned_word_end
C This loop runs at 8 cycles per iteration. It has been
C observed running at only 7 cycles, for this speed, the loop
C started at offset 0x2ac in the object file.
C FIXME: consider software pipelining, similarly to the memxor
C loop.
.Lmemxor3_aligned_word_loop:
ldmdb AP!, {r4,r5,r6}
ldmdb BP!, {r7,r8,r10}
subs N, #12
eor r4, r7
eor r5, r8
eor r6, r10
stmdb DST!, {r4, r5,r6}
bcs .Lmemxor3_aligned_word_loop
.Lmemxor3_aligned_word_end:
C We have 0-11 bytes left to do, and N holds number of bytes -12.
adds N, #4
bcc .Lmemxor3_aligned_lt_8
C Do 8 bytes more, leftover is in N
ldmdb AP!, {r4, r5}
ldmdb BP!, {r6, r7}
eor r4, r6
eor r5, r7
stmdb DST!, {r4,r5}
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_aligned_lt_8:
adds N, #4
bcc .Lmemxor3_aligned_lt_4
ldr r4, [AP,#-4]!
ldr r5, [BP,#-4]!
eor r4, r5
str r4, [DST,#-4]!
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_aligned_lt_4:
adds N, #4
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_uu:
cmp ACNT, BCNT
bic AP, #3
bic BP, #3
rsb ATNC, ACNT, #32
bne .Lmemxor3_uud
C AP and BP are unaligned in the same way
ldr r4, [AP]
ldr r6, [BP]
eor r4, r6
tst N, #4
itet eq
moveq r5, r4
subne N, #4
beq .Lmemxor3_uu_odd
.Lmemxor3_uu_loop:
ldr r5, [AP, #-4]!
ldr r6, [BP, #-4]!
eor r5, r6
lsl r4, ATNC
eor r4, r4, r5, lsr ACNT
str r4, [DST, #-4]!
.Lmemxor3_uu_odd:
ldr r4, [AP, #-4]!
ldr r6, [BP, #-4]!
eor r4, r6
lsl r5, ATNC
eor r5, r5, r4, lsr ACNT
str r5, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_uu_loop
adds N, #8
beq .Lmemxor3_done
C Leftover bytes in a4, low end
ror r4, ACNT
.Lmemxor3_uu_leftover:
ror r4, #24
strb r4, [DST, #-1]!
subs N, #1
beq .Lmemxor3_done
subs ACNT, #8
bne .Lmemxor3_uu_leftover
b .Lmemxor3_bytes
.Lmemxor3_uud:
C Both AP and BP unaligned, and in different ways
rsb BTNC, BCNT, #32
ldr r4, [AP]
ldr r6, [BP]
tst N, #4
ittet eq
moveq r5, r4
moveq r7, r6
subne N, #4
beq .Lmemxor3_uud_odd
.Lmemxor3_uud_loop:
ldr r5, [AP, #-4]!
ldr r7, [BP, #-4]!
lsl r4, ATNC
eor r4, r4, r6, lsl BTNC
eor r4, r4, r5, lsr ACNT
eor r4, r4, r7, lsr BCNT
str r4, [DST, #-4]!
.Lmemxor3_uud_odd:
ldr r4, [AP, #-4]!
ldr r6, [BP, #-4]!
lsl r5, ATNC
eor r5, r5, r7, lsl BTNC
eor r5, r5, r4, lsr ACNT
eor r5, r5, r6, lsr BCNT
str r5, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_uud_loop
adds N, #8
beq .Lmemxor3_done
C FIXME: More clever left-over handling? For now, just adjust pointers.
add AP, AP, ACNT, lsr #3
add BP, BP, BCNT, lsr #3
b .Lmemxor3_bytes
EPILOGUE(nettle_memxor3)
C arm/memxor3.asm
ifelse(<
Copyright (C) 2013, 2015 Niels Möller
This file is part of GNU Nettle.
GNU Nettle is free software: you can redistribute it and/or
modify it under the terms of either:
* the GNU Lesser General Public License as published by the Free
Software Foundation; either version 3 of the License, or (at your
option) any later version.
or
* the GNU General Public License as published by the Free
Software Foundation; either version 2 of the License, or (at your
option) any later version.
or both in parallel, as here.
GNU Nettle 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
General Public License for more details.
You should have received copies of the GNU General Public License and
the GNU Lesser General Public License along with this program. If
not, see http://www.gnu.org/licenses/.
>)
C Possible speedups:
C
C The ldm instruction can do load two registers per cycle,
C if the address is two-word aligned. Or three registers in two
C cycles, regardless of alignment.
C Register usage:
define(<DST>, <r0>)
define(<AP>, <r1>)
define(<BP>, <r2>)
define(<N>, <r3>)
C Temporaries r4-r7
define(<ACNT>, <r8>)
define(<ATNC>, <r10>)
define(<BCNT>, <r11>)
define(<BTNC>, <r12>)
.syntax unified
.file "memxor3.asm"
.text
.arm
C memxor3(void *dst, const void *a, const void *b, size_t n)
.align 2
PROLOGUE(nettle_memxor3)
cmp N, #0
beq .Lmemxor3_ret
push {r4,r5,r6,r7,r8,r10,r11}
cmp N, #7
add AP, N
add BP, N
add DST, N
bcs .Lmemxor3_large
C Simple byte loop
.Lmemxor3_bytes:
ldrb r4, [AP, #-1]!
ldrb r5, [BP, #-1]!
eor r4, r5
strb r4, [DST, #-1]!
subs N, #1
bne .Lmemxor3_bytes
.Lmemxor3_done:
pop {r4,r5,r6,r7,r8,r10,r11}
.Lmemxor3_ret:
bx lr
.Lmemxor3_align_loop:
ldrb r4, [AP, #-1]!
ldrb r5, [BP, #-1]!
eor r5, r4
strb r5, [DST, #-1]!
sub N, #1
.Lmemxor3_large:
tst DST, #3
bne .Lmemxor3_align_loop
C We have at least 4 bytes left to do here.
sub N, #4
ands ACNT, AP, #3
lsl ACNT, #3
beq .Lmemxor3_a_aligned
ands BCNT, BP, #3
lsl BCNT, #3
bne .Lmemxor3_uu
C Swap
mov r4, AP
mov AP, BP
mov BP, r4
.Lmemxor3_au:
C NOTE: We have the relevant shift count in ACNT, not BCNT
C AP is aligned, BP is not
C v original SRC
C +-------+------+
C |SRC-4 |SRC |
C +---+---+------+
C |DST-4 |
C +-------+
C
C With little-endian, we need to do
C DST[i-i] ^= (SRC[i-i] >> CNT) ^ (SRC[i] << TNC)
rsb ATNC, ACNT, #32
bic BP, #3
ldr r4, [BP]
tst N, #4
itet eq
moveq r5, r4
subne N, #4
beq .Lmemxor3_au_odd
.Lmemxor3_au_loop:
ldr r5, [BP, #-4]!
ldr r6, [AP, #-4]!
eor r6, r6, r4, lsl ATNC
eor r6, r6, r5, lsr ACNT
str r6, [DST, #-4]!
.Lmemxor3_au_odd:
ldr r4, [BP, #-4]!
ldr r6, [AP, #-4]!
eor r6, r6, r5, lsl ATNC
eor r6, r6, r4, lsr ACNT
str r6, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_au_loop
adds N, #8
beq .Lmemxor3_done
C Leftover bytes in r4, low end
ldr r5, [AP, #-4]
eor r4, r5, r4, lsl ATNC
.Lmemxor3_au_leftover:
C Store a byte at a time
ror r4, #24
strb r4, [DST, #-1]!
subs N, #1
beq .Lmemxor3_done
subs ACNT, #8
sub AP, #1
bne .Lmemxor3_au_leftover
b .Lmemxor3_bytes
.Lmemxor3_a_aligned:
ands ACNT, BP, #3
lsl ACNT, #3
bne .Lmemxor3_au ;
C a, b and dst all have the same alignment.
subs N, #8
bcc .Lmemxor3_aligned_word_end
C This loop runs at 8 cycles per iteration. It has been
C observed running at only 7 cycles, for this speed, the loop
C started at offset 0x2ac in the object file.
C FIXME: consider software pipelining, similarly to the memxor
C loop.
.Lmemxor3_aligned_word_loop:
ldmdb AP!, {r4,r5,r6}
ldmdb BP!, {r7,r8,r10}
subs N, #12
eor r4, r7
eor r5, r8
eor r6, r10
stmdb DST!, {r4, r5,r6}
bcs .Lmemxor3_aligned_word_loop
.Lmemxor3_aligned_word_end:
C We have 0-11 bytes left to do, and N holds number of bytes -12.
adds N, #4
bcc .Lmemxor3_aligned_lt_8
C Do 8 bytes more, leftover is in N
ldmdb AP!, {r4, r5}
ldmdb BP!, {r6, r7}
eor r4, r6
eor r5, r7
stmdb DST!, {r4,r5}
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_aligned_lt_8:
adds N, #4
bcc .Lmemxor3_aligned_lt_4
ldr r4, [AP,#-4]!
ldr r5, [BP,#-4]!
eor r4, r5
str r4, [DST,#-4]!
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_aligned_lt_4:
adds N, #4
beq .Lmemxor3_done
b .Lmemxor3_bytes
.Lmemxor3_uu:
cmp ACNT, BCNT
bic AP, #3
bic BP, #3
rsb ATNC, ACNT, #32
bne .Lmemxor3_uud
C AP and BP are unaligned in the same way
ldr r4, [AP]
ldr r6, [BP]
eor r4, r6
tst N, #4
itet eq
moveq r5, r4
subne N, #4
beq .Lmemxor3_uu_odd
.Lmemxor3_uu_loop:
ldr r5, [AP, #-4]!
ldr r6, [BP, #-4]!
eor r5, r6
lsl r4, ATNC
eor r4, r4, r5, lsr ACNT
str r4, [DST, #-4]!
.Lmemxor3_uu_odd:
ldr r4, [AP, #-4]!
ldr r6, [BP, #-4]!
eor r4, r6
lsl r5, ATNC
eor r5, r5, r4, lsr ACNT
str r5, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_uu_loop
adds N, #8
beq .Lmemxor3_done
C Leftover bytes in a4, low end
ror r4, ACNT
.Lmemxor3_uu_leftover:
ror r4, #24
strb r4, [DST, #-1]!
subs N, #1
beq .Lmemxor3_done
subs ACNT, #8
bne .Lmemxor3_uu_leftover
b .Lmemxor3_bytes
.Lmemxor3_uud:
C Both AP and BP unaligned, and in different ways
rsb BTNC, BCNT, #32
ldr r4, [AP]
ldr r6, [BP]
tst N, #4
ittet eq
moveq r5, r4
moveq r7, r6
subne N, #4
beq .Lmemxor3_uud_odd
.Lmemxor3_uud_loop:
ldr r5, [AP, #-4]!
ldr r7, [BP, #-4]!
lsl r4, ATNC
eor r4, r4, r6, lsl BTNC
eor r4, r4, r5, lsr ACNT
eor r4, r4, r7, lsr BCNT
str r4, [DST, #-4]!
.Lmemxor3_uud_odd:
ldr r4, [AP, #-4]!
ldr r6, [BP, #-4]!
lsl r5, ATNC
eor r5, r5, r7, lsl BTNC
eor r5, r5, r4, lsr ACNT
eor r5, r5, r6, lsr BCNT
str r5, [DST, #-4]!
subs N, #8
bcs .Lmemxor3_uud_loop
adds N, #8
beq .Lmemxor3_done
C FIXME: More clever left-over handling? For now, just adjust pointers.
add AP, AP, ACNT, lsr #3
add BP, BP, BCNT, lsr #3
b .Lmemxor3_bytes
EPILOGUE(nettle_memxor3)
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