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#include <asm/export.h> 47/* Allow an exception for an insn; exit if we get one. */ 48#define EX(x,y...) \ 49 99: x,##y; \ 50 .section __ex_table,"a"; \ 51 .long 99b - .; \ 52 lda $31, $exception-99b($31); \ 53 .previous 54 55 .set noat 56 .set noreorder 57 .align 4 58 59 .globl __do_clear_user 60 .ent __do_clear_user 61 .frame $30, 0, $28 62 .prologue 0 63 64 # Pipeline info : Slotting & Comments 65__do_clear_user: 66 and $6, 7, $4 # .. E .. .. : find dest head misalignment 67 beq $0, $zerolength # U .. .. .. : U L U L 68 69 addq $0, $4, $1 # .. .. .. E : bias counter 70 and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail 71# Note - we never actually use $2, so this is a moot computation 72# and we can rewrite this later... 73 srl $1, 3, $1 # .. E .. .. : number of quadwords to clear 74 beq $4, $headalign # U .. .. .. : U L U L 75 76/* 77 * Head is not aligned. Write (8 - $4) bytes to head of destination 78 * This means $6 is known to be misaligned 79 */ 80 EX( ldq_u $5, 0($6) ) # .. .. .. L : load dst word to mask back in 81 beq $1, $onebyte # .. .. U .. : sub-word store? 82 mskql $5, $6, $5 # .. U .. .. : take care of misaligned head 83 addq $6, 8, $6 # E .. .. .. : L U U L 84 85 EX( stq_u $5, -8($6) ) # .. .. .. L : 86 subq $1, 1, $1 # .. .. E .. : 87 addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment 88 subq $0, 8, $0 # E .. .. .. : U L U L 89 90 .align 4 91/* 92 * (The .align directive ought to be a moot point) 93 * values upon initial entry to the loop 94 * $1 is number of quadwords to clear (zero is a valid value) 95 * $2 is number of trailing bytes (0..7) ($2 never used...) 96 * $6 is known to be aligned 0mod8 97 */ 98$headalign: 99 subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop 100 and $6, 0x3f, $2 # .. .. E .. : Forward work for huge loop 101 subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) 102 blt $4, $trailquad # U .. .. .. : U L U L 103 104/* 105 * We know that we're going to do at least 16 quads, which means we are 106 * going to be able to use the large block clear loop at least once. 107 * Figure out how many quads we need to clear before we are 0mod64 aligned 108 * so we can use the wh64 instruction. 109 */ 110 111 nop # .. .. .. E 112 nop # .. .. E .. 113 nop # .. E .. .. 114 beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 115 116$alignmod64: 117 EX( stq_u $31, 0($6) ) # .. .. .. L 118 addq $3, 8, $3 # .. .. E .. 119 subq $0, 8, $0 # .. E .. .. 120 nop # E .. .. .. : U L U L 121 122 nop # .. .. .. E 123 subq $1, 1, $1 # .. .. E .. 124 addq $6, 8, $6 # .. E .. .. 125 blt $3, $alignmod64 # U .. .. .. : U L U L 126 127$bigalign: 128/* 129 * $0 is the number of bytes left 130 * $1 is the number of quads left 131 * $6 is aligned 0mod64 132 * we know that we'll be taking a minimum of one trip through 133 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle 134 * We are _not_ going to update $0 after every single store. That 135 * would be silly, because there will be cross-cluster dependencies 136 * no matter how the code is scheduled. By doing it in slightly 137 * staggered fashion, we can still do this loop in 5 fetches 138 * The worse case will be doing two extra quads in some future execution, 139 * in the event of an interrupted clear. 140 * Assumes the wh64 needs to be for 2 trips through the loop in the future 141 * The wh64 is issued on for the starting destination address for trip +2 142 * through the loop, and if there are less than two trips left, the target 143 * address will be for the current trip. 144 */ 145 nop # E : 146 nop # E : 147 nop # E : 148 bis $6,$6,$3 # E : U L U L : Initial wh64 address is dest 149 /* This might actually help for the current trip... */ 150 151$do_wh64: 152 wh64 ($3) # .. .. .. L1 : memory subsystem hint 153 subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? 154 EX( stq_u $31, 0($6) ) # .. L .. .. 155 subq $0, 8, $0 # E .. .. .. : U L U L 156 157 addq $6, 128, $3 # E : Target address of wh64 158 EX( stq_u $31, 8($6) ) # L : 159 EX( stq_u $31, 16($6) ) # L : 160 subq $0, 16, $0 # E : U L L U 161 162 nop # E : 163 EX( stq_u $31, 24($6) ) # L : 164 EX( stq_u $31, 32($6) ) # L : 165 subq $0, 168, $5 # E : U L L U : two trips through the loop left? 166 /* 168 = 192 - 24, since we've already completed some stores */ 167 168 subq $0, 16, $0 # E : 169 EX( stq_u $31, 40($6) ) # L : 170 EX( stq_u $31, 48($6) ) # L : 171 cmovlt $5, $6, $3 # E : U L L U : Latency 2, extra mapping cycle 172 173 subq $1, 8, $1 # E : 174 subq $0, 16, $0 # E : 175 EX( stq_u $31, 56($6) ) # L : 176 nop # E : U L U L 177 178 nop # E : 179 subq $0, 8, $0 # E : 180 addq $6, 64, $6 # E : 181 bge $4, $do_wh64 # U : U L U L 182 183$trailquad: 184 # zero to 16 quadwords left to store, plus any trailing bytes 185 # $1 is the number of quadwords left to go. 186 # 187 nop # .. .. .. E 188 nop # .. .. E .. 189 nop # .. E .. .. 190 beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go 191 192$onequad: 193 EX( stq_u $31, 0($6) ) # .. .. .. L 194 subq $1, 1, $1 # .. .. E .. 195 subq $0, 8, $0 # .. E .. .. 196 nop # E .. .. .. : U L U L 197 198 nop # .. .. .. E 199 nop # .. .. E .. 200 addq $6, 8, $6 # .. E .. .. 201 bgt $1, $onequad # U .. .. .. : U L U L 202 203 # We have an unknown number of bytes left to go. 204$trailbytes: 205 nop # .. .. .. E 206 nop # .. .. E .. 207 nop # .. E .. .. 208 beq $0, $zerolength # U .. .. .. : U L U L 209 210 # $0 contains the number of bytes left to copy (0..31) 211 # so we will use $0 as the loop counter 212 # We know for a fact that $0 > 0 zero due to previous context 213$onebyte: 214 EX( stb $31, 0($6) ) # .. .. .. L 215 subq $0, 1, $0 # .. .. E .. : 216 addq $6, 1, $6 # .. E .. .. : 217 bgt $0, $onebyte # U .. .. .. : U L U L 218 219$zerolength: 220$exception: # Destination for exception recovery(?) 221 nop # .. .. .. E : 222 nop # .. .. E .. : 223 nop # .. E .. .. : 224 ret $31, ($28), 1 # L0 .. .. .. : L U L U 225 .end __do_clear_user 226 EXPORT_SYMBOL(__do_clear_user) 227