ev6-memchr.S (5444B)
1/* SPDX-License-Identifier: GPL-2.0 */ 2/* 3 * arch/alpha/lib/ev6-memchr.S 4 * 5 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> 6 * 7 * Finds characters in a memory area. Optimized for the Alpha: 8 * 9 * - memory accessed as aligned quadwords only 10 * - uses cmpbge to compare 8 bytes in parallel 11 * - does binary search to find 0 byte in last 12 * quadword (HAKMEM needed 12 instructions to 13 * do this instead of the 9 instructions that 14 * binary search needs). 15 * 16 * For correctness consider that: 17 * 18 * - only minimum number of quadwords may be accessed 19 * - the third argument is an unsigned long 20 * 21 * Much of the information about 21264 scheduling/coding comes from: 22 * Compiler Writer's Guide for the Alpha 21264 23 * abbreviated as 'CWG' in other comments here 24 * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html 25 * Scheduling notation: 26 * E - either cluster 27 * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 28 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 29 * Try not to change the actual algorithm if possible for consistency. 30 */ 31#include <asm/export.h> 32 .set noreorder 33 .set noat 34 35 .align 4 36 .globl memchr 37 .ent memchr 38memchr: 39 .frame $30,0,$26,0 40 .prologue 0 41 42 # Hack -- if someone passes in (size_t)-1, hoping to just 43 # search til the end of the address space, we will overflow 44 # below when we find the address of the last byte. Given 45 # that we will never have a 56-bit address space, cropping 46 # the length is the easiest way to avoid trouble. 47 zap $18, 0x80, $5 # U : Bound length 48 beq $18, $not_found # U : 49 ldq_u $1, 0($16) # L : load first quadword Latency=3 50 and $17, 0xff, $17 # E : L L U U : 00000000000000ch 51 52 insbl $17, 1, $2 # U : 000000000000ch00 53 cmpult $18, 9, $4 # E : small (< 1 quad) string? 54 or $2, $17, $17 # E : 000000000000chch 55 lda $3, -1($31) # E : U L L U 56 57 sll $17, 16, $2 # U : 00000000chch0000 58 addq $16, $5, $5 # E : Max search address 59 or $2, $17, $17 # E : 00000000chchchch 60 sll $17, 32, $2 # U : U L L U : chchchch00000000 61 62 or $2, $17, $17 # E : chchchchchchchch 63 extql $1, $16, $7 # U : $7 is upper bits 64 beq $4, $first_quad # U : 65 ldq_u $6, -1($5) # L : L U U L : eight or less bytes to search Latency=3 66 67 extqh $6, $16, $6 # U : 2 cycle stall for $6 68 mov $16, $0 # E : 69 nop # E : 70 or $7, $6, $1 # E : L U L U $1 = quadword starting at $16 71 72 # Deal with the case where at most 8 bytes remain to be searched 73 # in $1. E.g.: 74 # $18 = 6 75 # $1 = ????c6c5c4c3c2c1 76$last_quad: 77 negq $18, $6 # E : 78 xor $17, $1, $1 # E : 79 srl $3, $6, $6 # U : $6 = mask of $18 bits set 80 cmpbge $31, $1, $2 # E : L U L U 81 82 nop 83 nop 84 and $2, $6, $2 # E : 85 beq $2, $not_found # U : U L U L 86 87$found_it: 88#ifdef CONFIG_ALPHA_EV67 89 /* 90 * Since we are guaranteed to have set one of the bits, we don't 91 * have to worry about coming back with a 0x40 out of cttz... 92 */ 93 cttz $2, $3 # U0 : 94 addq $0, $3, $0 # E : All done 95 nop # E : 96 ret # L0 : L U L U 97#else 98 /* 99 * Slow and clunky. It can probably be improved. 100 * An exercise left for others. 101 */ 102 negq $2, $3 # E : 103 and $2, $3, $2 # E : 104 and $2, 0x0f, $1 # E : 105 addq $0, 4, $3 # E : 106 107 cmoveq $1, $3, $0 # E : Latency 2, extra map cycle 108 nop # E : keep with cmov 109 and $2, 0x33, $1 # E : 110 addq $0, 2, $3 # E : U L U L : 2 cycle stall on $0 111 112 cmoveq $1, $3, $0 # E : Latency 2, extra map cycle 113 nop # E : keep with cmov 114 and $2, 0x55, $1 # E : 115 addq $0, 1, $3 # E : U L U L : 2 cycle stall on $0 116 117 cmoveq $1, $3, $0 # E : Latency 2, extra map cycle 118 nop 119 nop 120 ret # L0 : L U L U 121#endif 122 123 # Deal with the case where $18 > 8 bytes remain to be 124 # searched. $16 may not be aligned. 125 .align 4 126$first_quad: 127 andnot $16, 0x7, $0 # E : 128 insqh $3, $16, $2 # U : $2 = 0000ffffffffffff ($16<0:2> ff) 129 xor $1, $17, $1 # E : 130 or $1, $2, $1 # E : U L U L $1 = ====ffffffffffff 131 132 cmpbge $31, $1, $2 # E : 133 bne $2, $found_it # U : 134 # At least one byte left to process. 135 ldq $1, 8($0) # L : 136 subq $5, 1, $18 # E : U L U L 137 138 addq $0, 8, $0 # E : 139 # Make $18 point to last quad to be accessed (the 140 # last quad may or may not be partial). 141 andnot $18, 0x7, $18 # E : 142 cmpult $0, $18, $2 # E : 143 beq $2, $final # U : U L U L 144 145 # At least two quads remain to be accessed. 146 147 subq $18, $0, $4 # E : $4 <- nr quads to be processed 148 and $4, 8, $4 # E : odd number of quads? 149 bne $4, $odd_quad_count # U : 150 # At least three quads remain to be accessed 151 mov $1, $4 # E : L U L U : move prefetched value to correct reg 152 153 .align 4 154$unrolled_loop: 155 ldq $1, 8($0) # L : prefetch $1 156 xor $17, $4, $2 # E : 157 cmpbge $31, $2, $2 # E : 158 bne $2, $found_it # U : U L U L 159 160 addq $0, 8, $0 # E : 161 nop # E : 162 nop # E : 163 nop # E : 164 165$odd_quad_count: 166 xor $17, $1, $2 # E : 167 ldq $4, 8($0) # L : prefetch $4 168 cmpbge $31, $2, $2 # E : 169 addq $0, 8, $6 # E : 170 171 bne $2, $found_it # U : 172 cmpult $6, $18, $6 # E : 173 addq $0, 8, $0 # E : 174 nop # E : 175 176 bne $6, $unrolled_loop # U : 177 mov $4, $1 # E : move prefetched value into $1 178 nop # E : 179 nop # E : 180 181$final: subq $5, $0, $18 # E : $18 <- number of bytes left to do 182 nop # E : 183 nop # E : 184 bne $18, $last_quad # U : 185 186$not_found: 187 mov $31, $0 # E : 188 nop # E : 189 nop # E : 190 ret # L0 : 191 192 .end memchr 193 EXPORT_SYMBOL(memchr)