1/* 2 * arch/alpha/lib/ev6-clear_user.S 3 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> 4 * 5 * Zero user space, handling exceptions as we go. 6 * 7 * We have to make sure that $0 is always up-to-date and contains the 8 * right "bytes left to zero" value (and that it is updated only _after_ 9 * a successful copy). There is also some rather minor exception setup 10 * stuff. 11 * 12 * NOTE! This is not directly C-callable, because the calling semantics 13 * are different: 14 * 15 * Inputs: 16 * length in $0 17 * destination address in $6 18 * exception pointer in $7 19 * return address in $28 (exceptions expect it there) 20 * 21 * Outputs: 22 * bytes left to copy in $0 23 * 24 * Clobbers: 25 * $1,$2,$3,$4,$5,$6 26 * 27 * Much of the information about 21264 scheduling/coding comes from: 28 * Compiler Writer's Guide for the Alpha 21264 29 * abbreviated as 'CWG' in other comments here 30 * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html 31 * Scheduling notation: 32 * E - either cluster 33 * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 34 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 35 * Try not to change the actual algorithm if possible for consistency. 36 * Determining actual stalls (other than slotting) doesn't appear to be easy to do. 37 * From perusing the source code context where this routine is called, it is 38 * a fair assumption that significant fractions of entire pages are zeroed, so 39 * it's going to be worth the effort to hand-unroll a big loop, and use wh64. 40 * ASSUMPTION: 41 * The believed purpose of only updating $0 after a store is that a signal 42 * may come along during the execution of this chunk of code, and we don't 43 * want to leave a hole (and we also want to avoid repeating lots of work) 44 */ 45 46/* Allow an exception for an insn; exit if we get one. */ 47#define EX(x,y...) \ 48 99: x,##y; \ 49 .section __ex_table,"a"; \ 50 .long 99b - .; \ 51 lda $31, $exception-99b($31); \ 52 .previous 53 54 .set noat 55 .set noreorder 56 .align 4 57 58 .globl __do_clear_user 59 .ent __do_clear_user 60 .frame $30, 0, $28 61 .prologue 0 62 63 # Pipeline info : Slotting & Comments 64__do_clear_user: 65 and $6, 7, $4 # .. E .. .. : find dest head misalignment 66 beq $0, $zerolength # U .. .. .. : U L U L 67 68 addq $0, $4, $1 # .. .. .. E : bias counter 69 and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail 70# Note - we never actually use $2, so this is a moot computation 71# and we can rewrite this later... 72 srl $1, 3, $1 # .. E .. .. : number of quadwords to clear 73 beq $4, $headalign # U .. .. .. : U L U L 74 75/* 76 * Head is not aligned. Write (8 - $4) bytes to head of destination 77 * This means $6 is known to be misaligned 78 */ 79 EX( ldq_u $5, 0($6) ) # .. .. .. L : load dst word to mask back in 80 beq $1, $onebyte # .. .. U .. : sub-word store? 81 mskql $5, $6, $5 # .. U .. .. : take care of misaligned head 82 addq $6, 8, $6 # E .. .. .. : L U U L 83 84 EX( stq_u $5, -8($6) ) # .. .. .. L : 85 subq $1, 1, $1 # .. .. E .. : 86 addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment 87 subq $0, 8, $0 # E .. .. .. : U L U L 88 89 .align 4 90/* 91 * (The .align directive ought to be a moot point) 92 * values upon initial entry to the loop 93 * $1 is number of quadwords to clear (zero is a valid value) 94 * $2 is number of trailing bytes (0..7) ($2 never used...) 95 * $6 is known to be aligned 0mod8 96 */ 97$headalign: 98 subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop 99 and $6, 0x3f, $2 # .. .. E .. : Forward work for huge loop 100 subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) 101 blt $4, $trailquad # U .. .. .. : U L U L 102 103/* 104 * We know that we're going to do at least 16 quads, which means we are 105 * going to be able to use the large block clear loop at least once. 106 * Figure out how many quads we need to clear before we are 0mod64 aligned 107 * so we can use the wh64 instruction. 108 */ 109 110 nop # .. .. .. E 111 nop # .. .. E .. 112 nop # .. E .. .. 113 beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 114 115$alignmod64: 116 EX( stq_u $31, 0($6) ) # .. .. .. L 117 addq $3, 8, $3 # .. .. E .. 118 subq $0, 8, $0 # .. E .. .. 119 nop # E .. .. .. : U L U L 120 121 nop # .. .. .. E 122 subq $1, 1, $1 # .. .. E .. 123 addq $6, 8, $6 # .. E .. .. 124 blt $3, $alignmod64 # U .. .. .. : U L U L 125 126$bigalign: 127/* 128 * $0 is the number of bytes left 129 * $1 is the number of quads left 130 * $6 is aligned 0mod64 131 * we know that we'll be taking a minimum of one trip through 132 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle 133 * We are _not_ going to update $0 after every single store. That 134 * would be silly, because there will be cross-cluster dependencies 135 * no matter how the code is scheduled. By doing it in slightly 136 * staggered fashion, we can still do this loop in 5 fetches 137 * The worse case will be doing two extra quads in some future execution, 138 * in the event of an interrupted clear. 139 * Assumes the wh64 needs to be for 2 trips through the loop in the future 140 * The wh64 is issued on for the starting destination address for trip +2 141 * through the loop, and if there are less than two trips left, the target 142 * address will be for the current trip. 143 */ 144 nop # E : 145 nop # E : 146 nop # E : 147 bis $6,$6,$3 # E : U L U L : Initial wh64 address is dest 148 /* This might actually help for the current trip... */ 149 150$do_wh64: 151 wh64 ($3) # .. .. .. L1 : memory subsystem hint 152 subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? 153 EX( stq_u $31, 0($6) ) # .. L .. .. 154 subq $0, 8, $0 # E .. .. .. : U L U L 155 156 addq $6, 128, $3 # E : Target address of wh64 157 EX( stq_u $31, 8($6) ) # L : 158 EX( stq_u $31, 16($6) ) # L : 159 subq $0, 16, $0 # E : U L L U 160 161 nop # E : 162 EX( stq_u $31, 24($6) ) # L : 163 EX( stq_u $31, 32($6) ) # L : 164 subq $0, 168, $5 # E : U L L U : two trips through the loop left? 165 /* 168 = 192 - 24, since we've already completed some stores */ 166 167 subq $0, 16, $0 # E : 168 EX( stq_u $31, 40($6) ) # L : 169 EX( stq_u $31, 48($6) ) # L : 170 cmovlt $5, $6, $3 # E : U L L U : Latency 2, extra mapping cycle 171 172 subq $1, 8, $1 # E : 173 subq $0, 16, $0 # E : 174 EX( stq_u $31, 56($6) ) # L : 175 nop # E : U L U L 176 177 nop # E : 178 subq $0, 8, $0 # E : 179 addq $6, 64, $6 # E : 180 bge $4, $do_wh64 # U : U L U L 181 182$trailquad: 183 # zero to 16 quadwords left to store, plus any trailing bytes 184 # $1 is the number of quadwords left to go. 185 # 186 nop # .. .. .. E 187 nop # .. .. E .. 188 nop # .. E .. .. 189 beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go 190 191$onequad: 192 EX( stq_u $31, 0($6) ) # .. .. .. L 193 subq $1, 1, $1 # .. .. E .. 194 subq $0, 8, $0 # .. E .. .. 195 nop # E .. .. .. : U L U L 196 197 nop # .. .. .. E 198 nop # .. .. E .. 199 addq $6, 8, $6 # .. E .. .. 200 bgt $1, $onequad # U .. .. .. : U L U L 201 202 # We have an unknown number of bytes left to go. 203$trailbytes: 204 nop # .. .. .. E 205 nop # .. .. E .. 206 nop # .. E .. .. 207 beq $0, $zerolength # U .. .. .. : U L U L 208 209 # $0 contains the number of bytes left to copy (0..31) 210 # so we will use $0 as the loop counter 211 # We know for a fact that $0 > 0 zero due to previous context 212$onebyte: 213 EX( stb $31, 0($6) ) # .. .. .. L 214 subq $0, 1, $0 # .. .. E .. : 215 addq $6, 1, $6 # .. E .. .. : 216 bgt $0, $onebyte # U .. .. .. : U L U L 217 218$zerolength: 219$exception: # Destination for exception recovery(?) 220 nop # .. .. .. E : 221 nop # .. .. E .. : 222 nop # .. E .. .. : 223 ret $31, ($28), 1 # L0 .. .. .. : L U L U 224 .end __do_clear_user 225