ev6-copy_user.S (7016B)
1/* SPDX-License-Identifier: GPL-2.0 */ 2/* 3 * arch/alpha/lib/ev6-copy_user.S 4 * 5 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> 6 * 7 * Copy to/from user space, handling exceptions as we go.. This 8 * isn't exactly pretty. 9 * 10 * This is essentially the same as "memcpy()", but with a few twists. 11 * Notably, we have to make sure that $0 is always up-to-date and 12 * contains the right "bytes left to copy" value (and that it is updated 13 * only _after_ a successful copy). There is also some rather minor 14 * exception setup stuff.. 15 * 16 * Much of the information about 21264 scheduling/coding comes from: 17 * Compiler Writer's Guide for the Alpha 21264 18 * abbreviated as 'CWG' in other comments here 19 * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html 20 * Scheduling notation: 21 * E - either cluster 22 * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 23 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 24 */ 25 26#include <asm/export.h> 27/* Allow an exception for an insn; exit if we get one. */ 28#define EXI(x,y...) \ 29 99: x,##y; \ 30 .section __ex_table,"a"; \ 31 .long 99b - .; \ 32 lda $31, $exitin-99b($31); \ 33 .previous 34 35#define EXO(x,y...) \ 36 99: x,##y; \ 37 .section __ex_table,"a"; \ 38 .long 99b - .; \ 39 lda $31, $exitout-99b($31); \ 40 .previous 41 42 .set noat 43 .align 4 44 .globl __copy_user 45 .ent __copy_user 46 # Pipeline info: Slotting & Comments 47__copy_user: 48 .prologue 0 49 mov $18, $0 # .. .. .. E 50 subq $18, 32, $1 # .. .. E. .. : Is this going to be a small copy? 51 nop # .. E .. .. 52 beq $18, $zerolength # U .. .. .. : U L U L 53 54 and $16,7,$3 # .. .. .. E : is leading dest misalignment 55 ble $1, $onebyteloop # .. .. U .. : 1st branch : small amount of data 56 beq $3, $destaligned # .. U .. .. : 2nd (one cycle fetcher stall) 57 subq $3, 8, $3 # E .. .. .. : L U U L : trip counter 58/* 59 * The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U) 60 * This loop aligns the destination a byte at a time 61 * We know we have at least one trip through this loop 62 */ 63$aligndest: 64 EXI( ldbu $1,0($17) ) # .. .. .. L : Keep loads separate from stores 65 addq $16,1,$16 # .. .. E .. : Section 3.8 in the CWG 66 addq $3,1,$3 # .. E .. .. : 67 nop # E .. .. .. : U L U L 68 69/* 70 * the -1 is to compensate for the inc($16) done in a previous quadpack 71 * which allows us zero dependencies within either quadpack in the loop 72 */ 73 EXO( stb $1,-1($16) ) # .. .. .. L : 74 addq $17,1,$17 # .. .. E .. : Section 3.8 in the CWG 75 subq $0,1,$0 # .. E .. .. : 76 bne $3, $aligndest # U .. .. .. : U L U L 77 78/* 79 * If we fell through into here, we have a minimum of 33 - 7 bytes 80 * If we arrived via branch, we have a minimum of 32 bytes 81 */ 82$destaligned: 83 and $17,7,$1 # .. .. .. E : Check _current_ source alignment 84 bic $0,7,$4 # .. .. E .. : number bytes as a quadword loop 85 EXI( ldq_u $3,0($17) ) # .. L .. .. : Forward fetch for fallthrough code 86 beq $1,$quadaligned # U .. .. .. : U L U L 87 88/* 89 * In the worst case, we've just executed an ldq_u here from 0($17) 90 * and we'll repeat it once if we take the branch 91 */ 92 93/* Misaligned quadword loop - not unrolled. Leave it that way. */ 94$misquad: 95 EXI( ldq_u $2,8($17) ) # .. .. .. L : 96 subq $4,8,$4 # .. .. E .. : 97 extql $3,$17,$3 # .. U .. .. : 98 extqh $2,$17,$1 # U .. .. .. : U U L L 99 100 bis $3,$1,$1 # .. .. .. E : 101 EXO( stq $1,0($16) ) # .. .. L .. : 102 addq $17,8,$17 # .. E .. .. : 103 subq $0,8,$0 # E .. .. .. : U L L U 104 105 addq $16,8,$16 # .. .. .. E : 106 bis $2,$2,$3 # .. .. E .. : 107 nop # .. E .. .. : 108 bne $4,$misquad # U .. .. .. : U L U L 109 110 nop # .. .. .. E 111 nop # .. .. E .. 112 nop # .. E .. .. 113 beq $0,$zerolength # U .. .. .. : U L U L 114 115/* We know we have at least one trip through the byte loop */ 116 EXI ( ldbu $2,0($17) ) # .. .. .. L : No loads in the same quad 117 addq $16,1,$16 # .. .. E .. : as the store (Section 3.8 in CWG) 118 nop # .. E .. .. : 119 br $31, $dirtyentry # L0 .. .. .. : L U U L 120/* Do the trailing byte loop load, then hop into the store part of the loop */ 121 122/* 123 * A minimum of (33 - 7) bytes to do a quad at a time. 124 * Based upon the usage context, it's worth the effort to unroll this loop 125 * $0 - number of bytes to be moved 126 * $4 - number of bytes to move as quadwords 127 * $16 is current destination address 128 * $17 is current source address 129 */ 130$quadaligned: 131 subq $4, 32, $2 # .. .. .. E : do not unroll for small stuff 132 nop # .. .. E .. 133 nop # .. E .. .. 134 blt $2, $onequad # U .. .. .. : U L U L 135 136/* 137 * There is a significant assumption here that the source and destination 138 * addresses differ by more than 32 bytes. In this particular case, a 139 * sparsity of registers further bounds this to be a minimum of 8 bytes. 140 * But if this isn't met, then the output result will be incorrect. 141 * Furthermore, due to a lack of available registers, we really can't 142 * unroll this to be an 8x loop (which would enable us to use the wh64 143 * instruction memory hint instruction). 144 */ 145$unroll4: 146 EXI( ldq $1,0($17) ) # .. .. .. L 147 EXI( ldq $2,8($17) ) # .. .. L .. 148 subq $4,32,$4 # .. E .. .. 149 nop # E .. .. .. : U U L L 150 151 addq $17,16,$17 # .. .. .. E 152 EXO( stq $1,0($16) ) # .. .. L .. 153 EXO( stq $2,8($16) ) # .. L .. .. 154 subq $0,16,$0 # E .. .. .. : U L L U 155 156 addq $16,16,$16 # .. .. .. E 157 EXI( ldq $1,0($17) ) # .. .. L .. 158 EXI( ldq $2,8($17) ) # .. L .. .. 159 subq $4, 32, $3 # E .. .. .. : U U L L : is there enough for another trip? 160 161 EXO( stq $1,0($16) ) # .. .. .. L 162 EXO( stq $2,8($16) ) # .. .. L .. 163 subq $0,16,$0 # .. E .. .. 164 addq $17,16,$17 # E .. .. .. : U L L U 165 166 nop # .. .. .. E 167 nop # .. .. E .. 168 addq $16,16,$16 # .. E .. .. 169 bgt $3,$unroll4 # U .. .. .. : U L U L 170 171 nop 172 nop 173 nop 174 beq $4, $noquads 175 176$onequad: 177 EXI( ldq $1,0($17) ) 178 subq $4,8,$4 179 addq $17,8,$17 180 nop 181 182 EXO( stq $1,0($16) ) 183 subq $0,8,$0 184 addq $16,8,$16 185 bne $4,$onequad 186 187$noquads: 188 nop 189 nop 190 nop 191 beq $0,$zerolength 192 193/* 194 * For small copies (or the tail of a larger copy), do a very simple byte loop. 195 * There's no point in doing a lot of complex alignment calculations to try to 196 * to quadword stuff for a small amount of data. 197 * $0 - remaining number of bytes left to copy 198 * $16 - current dest addr 199 * $17 - current source addr 200 */ 201 202$onebyteloop: 203 EXI ( ldbu $2,0($17) ) # .. .. .. L : No loads in the same quad 204 addq $16,1,$16 # .. .. E .. : as the store (Section 3.8 in CWG) 205 nop # .. E .. .. : 206 nop # E .. .. .. : U L U L 207 208$dirtyentry: 209/* 210 * the -1 is to compensate for the inc($16) done in a previous quadpack 211 * which allows us zero dependencies within either quadpack in the loop 212 */ 213 EXO ( stb $2,-1($16) ) # .. .. .. L : 214 addq $17,1,$17 # .. .. E .. : quadpack as the load 215 subq $0,1,$0 # .. E .. .. : change count _after_ copy 216 bgt $0,$onebyteloop # U .. .. .. : U L U L 217 218$zerolength: 219$exitin: 220$exitout: # Destination for exception recovery(?) 221 nop # .. .. .. E 222 nop # .. .. E .. 223 nop # .. E .. .. 224 ret $31,($26),1 # L0 .. .. .. : L U L U 225 226 .end __copy_user 227 EXPORT_SYMBOL(__copy_user)