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