/* SPDX-License-Identifier: GPL-2.0 */
/*
* arch/alpha/lib/ev6-memchr.S
*
* 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
*
* Finds characters in a memory area. Optimized for the Alpha:
*
* - memory accessed as aligned quadwords only
* - uses cmpbge to compare 8 bytes in parallel
* - does binary search to find 0 byte in last
* quadword (HAKMEM needed 12 instructions to
* do this instead of the 9 instructions that
* binary search needs).
*
* For correctness consider that:
*
* - only minimum number of quadwords may be accessed
* - the third argument is an unsigned long
*
* Much of the information about 21264 scheduling/coding comes from:
* Compiler Writer's Guide for the Alpha 21264
* abbreviated as 'CWG' in other comments here
* ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
* Scheduling notation:
* E - either cluster
* U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1
* L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1
* Try not to change the actual algorithm if possible for consistency.
*/
#include <linux/export.h>
.set noreorder
.set noat
.align 4
.globl memchr
.ent memchr
memchr:
.frame $30,0,$26,0
.prologue 0
# Hack -- if someone passes in (size_t)-1, hoping to just
# search til the end of the address space, we will overflow
# below when we find the address of the last byte. Given
# that we will never have a 56-bit address space, cropping
# the length is the easiest way to avoid trouble.
zap $18, 0x80, $5 # U : Bound length
beq $18, $not_found # U :
ldq_u $1, 0($16) # L : load first quadword Latency=3
and $17, 0xff, $17 # E : L L U U : 00000000000000ch
insbl $17, 1, $2 # U : 000000000000ch00
cmpult $18, 9, $4 # E : small (< 1 quad) string?
or $2, $17, $17 # E : 000000000000chch
lda $3, -1($31) # E : U L L U
sll $17, 16, $2 # U : 00000000chch0000
addq $16, $5, $5 # E : Max search address
or $2, $17, $17 # E : 00000000chchchch
sll $17, 32, $2 # U : U L L U : chchchch00000000
or $2, $17, $17 # E : chchchchchchchch
extql $1, $16, $7 # U : $7 is upper bits
beq $4, $first_quad # U :
ldq_u $6, -1($5) # L : L U U L : eight or less bytes to search Latency=3
extqh $6, $16, $6 # U : 2 cycle stall for $6
mov $16, $0 # E :
nop # E :
or $7, $6, $1 # E : L U L U $1 = quadword starting at $16
# Deal with the case where at most 8 bytes remain to be searched
# in $1. E.g.:
# $18 = 6
# $1 = ????c6c5c4c3c2c1
$last_quad:
negq $18, $6 # E :
xor $17, $1, $1 # E :
srl $3, $6, $6 # U : $6 = mask of $18 bits set
cmpbge $31, $1, $2 # E : L U L U
nop
nop
and $2, $6, $2 # E :
beq $2, $not_found # U : U L U L
$found_it:
#ifdef CONFIG_ALPHA_EV67
/*
* Since we are guaranteed to have set one of the bits, we don't
* have to worry about coming back with a 0x40 out of cttz...
*/
cttz $2, $3 # U0 :
addq $0, $3, $0 # E : All done
nop # E :
ret # L0 : L U L U
#else
/*
* Slow and clunky. It can probably be improved.
* An exercise left for others.
*/
negq $2, $3 # E :
and $2, $3, $2 # E :
and $2, 0x0f, $1 # E :
addq $0, 4, $3 # E :
cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
nop # E : keep with cmov
and $2, 0x33, $1 # E :
addq $0, 2, $3 # E : U L U L : 2 cycle stall on $0
cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
nop # E : keep with cmov
and $2, 0x55, $1 # E :
addq $0, 1, $3 # E : U L U L : 2 cycle stall on $0
cmoveq $1, $3, $0 # E : Latency 2, extra map cycle
nop
nop
ret # L0 : L U L U
#endif
# Deal with the case where $18 > 8 bytes remain to be
# searched. $16 may not be aligned.
.align 4
$first_quad:
andnot $16, 0x7, $0 # E :
insqh $3, $16, $2 # U : $2 = 0000ffffffffffff ($16<0:2> ff)
xor $1, $17, $1 # E :
or $1, $2, $1 # E : U L U L $1 = ====ffffffffffff
cmpbge $31, $1, $2 # E :
bne $2, $found_it # U :
# At least one byte left to process.
ldq $1, 8($0) # L :
subq $5, 1, $18 # E : U L U L
addq $0, 8, $0 # E :
# Make $18 point to last quad to be accessed (the
# last quad may or may not be partial).
andnot $18, 0x7, $18 # E :
cmpult $0, $18, $2 # E :
beq $2, $final # U : U L U L
# At least two quads remain to be accessed.
subq $18, $0, $4 # E : $4 <- nr quads to be processed
and $4, 8, $4 # E : odd number of quads?
bne $4, $odd_quad_count # U :
# At least three quads remain to be accessed
mov $1, $4 # E : L U L U : move prefetched value to correct reg
.align 4
$unrolled_loop:
ldq $1, 8($0) # L : prefetch $1
xor $17, $4, $2 # E :
cmpbge $31, $2, $2 # E :
bne $2, $found_it # U : U L U L
addq $0, 8, $0 # E :
nop # E :
nop # E :
nop # E :
$odd_quad_count:
xor $17, $1, $2 # E :
ldq $4, 8($0) # L : prefetch $4
cmpbge $31, $2, $2 # E :
addq $0, 8, $6 # E :
bne $2, $found_it # U :
cmpult $6, $18, $6 # E :
addq $0, 8, $0 # E :
nop # E :
bne $6, $unrolled_loop # U :
mov $4, $1 # E : move prefetched value into $1
nop # E :
nop # E :
$final: subq $5, $0, $18 # E : $18 <- number of bytes left to do
nop # E :
nop # E :
bne $18, $last_quad # U :
$not_found:
mov $31, $0 # E :
nop # E :
nop # E :
ret # L0 :
.end memchr
EXPORT_SYMBOL(memchr)