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 .gprel32 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 ldgp $29,0($27) # we do exceptions -- we need the gp. 66 # Macro instruction becomes ldah/lda 67 # .. .. E E : 68 and $6, 7, $4 # .. E .. .. : find dest head misalignment 69 beq $0, $zerolength # U .. .. .. : U L U L 70 71 addq $0, $4, $1 # .. .. .. E : bias counter 72 and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail 73# Note - we never actually use $2, so this is a moot computation 74# and we can rewrite this later... 75 srl $1, 3, $1 # .. E .. .. : number of quadwords to clear 76 beq $4, $headalign # U .. .. .. : U L U L 77 78/* 79 * Head is not aligned. Write (8 - $4) bytes to head of destination 80 * This means $6 is known to be misaligned 81 */ 82 EX( ldq_u $5, 0($6) ) # .. .. .. L : load dst word to mask back in 83 beq $1, $onebyte # .. .. U .. : sub-word store? 84 mskql $5, $6, $5 # .. U .. .. : take care of misaligned head 85 addq $6, 8, $6 # E .. .. .. : L U U L 86 87 EX( stq_u $5, -8($6) ) # .. .. .. L : 88 subq $1, 1, $1 # .. .. E .. : 89 addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment 90 subq $0, 8, $0 # E .. .. .. : U L U L 91 92 .align 4 93/* 94 * (The .align directive ought to be a moot point) 95 * values upon initial entry to the loop 96 * $1 is number of quadwords to clear (zero is a valid value) 97 * $2 is number of trailing bytes (0..7) ($2 never used...) 98 * $6 is known to be aligned 0mod8 99 */ 100$headalign: 101 subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop 102 and $6, 0x3f, $2 # .. .. E .. : Forward work for huge loop 103 subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) 104 blt $4, $trailquad # U .. .. .. : U L U L 105 106/* 107 * We know that we're going to do at least 16 quads, which means we are 108 * going to be able to use the large block clear loop at least once. 109 * Figure out how many quads we need to clear before we are 0mod64 aligned 110 * so we can use the wh64 instruction. 111 */ 112 113 nop # .. .. .. E 114 nop # .. .. E .. 115 nop # .. E .. .. 116 beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 117 118$alignmod64: 119 EX( stq_u $31, 0($6) ) # .. .. .. L 120 addq $3, 8, $3 # .. .. E .. 121 subq $0, 8, $0 # .. E .. .. 122 nop # E .. .. .. : U L U L 123 124 nop # .. .. .. E 125 subq $1, 1, $1 # .. .. E .. 126 addq $6, 8, $6 # .. E .. .. 127 blt $3, $alignmod64 # U .. .. .. : U L U L 128 129$bigalign: 130/* 131 * $0 is the number of bytes left 132 * $1 is the number of quads left 133 * $6 is aligned 0mod64 134 * we know that we'll be taking a minimum of one trip through 135 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle 136 * We are _not_ going to update $0 after every single store. That 137 * would be silly, because there will be cross-cluster dependencies 138 * no matter how the code is scheduled. By doing it in slightly 139 * staggered fashion, we can still do this loop in 5 fetches 140 * The worse case will be doing two extra quads in some future execution, 141 * in the event of an interrupted clear. 142 * Assumes the wh64 needs to be for 2 trips through the loop in the future 143 * The wh64 is issued on for the starting destination address for trip +2 144 * through the loop, and if there are less than two trips left, the target 145 * address will be for the current trip. 146 */ 147 nop # E : 148 nop # E : 149 nop # E : 150 bis $6,$6,$3 # E : U L U L : Initial wh64 address is dest 151 /* This might actually help for the current trip... */ 152 153$do_wh64: 154 wh64 ($3) # .. .. .. L1 : memory subsystem hint 155 subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? 156 EX( stq_u $31, 0($6) ) # .. L .. .. 157 subq $0, 8, $0 # E .. .. .. : U L U L 158 159 addq $6, 128, $3 # E : Target address of wh64 160 EX( stq_u $31, 8($6) ) # L : 161 EX( stq_u $31, 16($6) ) # L : 162 subq $0, 16, $0 # E : U L L U 163 164 nop # E : 165 EX( stq_u $31, 24($6) ) # L : 166 EX( stq_u $31, 32($6) ) # L : 167 subq $0, 168, $5 # E : U L L U : two trips through the loop left? 168 /* 168 = 192 - 24, since we've already completed some stores */ 169 170 subq $0, 16, $0 # E : 171 EX( stq_u $31, 40($6) ) # L : 172 EX( stq_u $31, 48($6) ) # L : 173 cmovlt $5, $6, $3 # E : U L L U : Latency 2, extra mapping cycle 174 175 subq $1, 8, $1 # E : 176 subq $0, 16, $0 # E : 177 EX( stq_u $31, 56($6) ) # L : 178 nop # E : U L U L 179 180 nop # E : 181 subq $0, 8, $0 # E : 182 addq $6, 64, $6 # E : 183 bge $4, $do_wh64 # U : U L U L 184 185$trailquad: 186 # zero to 16 quadwords left to store, plus any trailing bytes 187 # $1 is the number of quadwords left to go. 188 # 189 nop # .. .. .. E 190 nop # .. .. E .. 191 nop # .. E .. .. 192 beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go 193 194$onequad: 195 EX( stq_u $31, 0($6) ) # .. .. .. L 196 subq $1, 1, $1 # .. .. E .. 197 subq $0, 8, $0 # .. E .. .. 198 nop # E .. .. .. : U L U L 199 200 nop # .. .. .. E 201 nop # .. .. E .. 202 addq $6, 8, $6 # .. E .. .. 203 bgt $1, $onequad # U .. .. .. : U L U L 204 205 # We have an unknown number of bytes left to go. 206$trailbytes: 207 nop # .. .. .. E 208 nop # .. .. E .. 209 nop # .. E .. .. 210 beq $0, $zerolength # U .. .. .. : U L U L 211 212 # $0 contains the number of bytes left to copy (0..31) 213 # so we will use $0 as the loop counter 214 # We know for a fact that $0 > 0 zero due to previous context 215$onebyte: 216 EX( stb $31, 0($6) ) # .. .. .. L 217 subq $0, 1, $0 # .. .. E .. : 218 addq $6, 1, $6 # .. E .. .. : 219 bgt $0, $onebyte # U .. .. .. : U L U L 220 221$zerolength: 222$exception: # Destination for exception recovery(?) 223 nop # .. .. .. E : 224 nop # .. .. E .. : 225 nop # .. E .. .. : 226 ret $31, ($28), 1 # L0 .. .. .. : L U L U 227 .end __do_clear_user 228 229