1 /* SPDX-License-Identifier: GPL-2.0 */ 2 /* 3 * arch/alpha/lib/ev6-clear_user.S 4 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> 5 * 6 * Zero user space, handling exceptions as we go. 7 * 8 * We have to make sure that $0 is always up-to-date and contains the 9 * right "bytes left to zero" value (and that it is updated only _after_ 10 * a successful copy). There is also some rather minor exception setup 11 * stuff. 12 * 13 * Much of the information about 21264 scheduling/coding comes from: 14 * Compiler Writer's Guide for the Alpha 21264 15 * abbreviated as 'CWG' in other comments here 16 * ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html 17 * Scheduling notation: 18 * E - either cluster 19 * U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 20 * L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 21 * Try not to change the actual algorithm if possible for consistency. 22 * Determining actual stalls (other than slotting) doesn't appear to be easy to do. 23 * From perusing the source code context where this routine is called, it is 24 * a fair assumption that significant fractions of entire pages are zeroed, so 25 * it's going to be worth the effort to hand-unroll a big loop, and use wh64. 26 * ASSUMPTION: 27 * The believed purpose of only updating $0 after a store is that a signal 28 * may come along during the execution of this chunk of code, and we don't 29 * want to leave a hole (and we also want to avoid repeating lots of work) 30 */ 31 32 #include <asm/export.h> 33 /* Allow an exception for an insn; exit if we get one. */ 34 #define EX(x,y...) \ 35 99: x,##y; \ 36 .section __ex_table,"a"; \ 37 .long 99b - .; \ 38 lda $31, $exception-99b($31); \ 39 .previous 40 41 .set noat 42 .set noreorder 43 .align 4 44 45 .globl __clear_user 46 .ent __clear_user 47 .frame $30, 0, $26 48 .prologue 0 49 50 # Pipeline info : Slotting & Comments 51 __clear_user: 52 and $17, $17, $0 53 and $16, 7, $4 # .. E .. .. : find dest head misalignment 54 beq $0, $zerolength # U .. .. .. : U L U L 55 56 addq $0, $4, $1 # .. .. .. E : bias counter 57 and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail 58 # Note - we never actually use $2, so this is a moot computation 59 # and we can rewrite this later... 60 srl $1, 3, $1 # .. E .. .. : number of quadwords to clear 61 beq $4, $headalign # U .. .. .. : U L U L 62 63 /* 64 * Head is not aligned. Write (8 - $4) bytes to head of destination 65 * This means $16 is known to be misaligned 66 */ 67 EX( ldq_u $5, 0($16) ) # .. .. .. L : load dst word to mask back in 68 beq $1, $onebyte # .. .. U .. : sub-word store? 69 mskql $5, $16, $5 # .. U .. .. : take care of misaligned head 70 addq $16, 8, $16 # E .. .. .. : L U U L 71 72 EX( stq_u $5, -8($16) ) # .. .. .. L : 73 subq $1, 1, $1 # .. .. E .. : 74 addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment 75 subq $0, 8, $0 # E .. .. .. : U L U L 76 77 .align 4 78 /* 79 * (The .align directive ought to be a moot point) 80 * values upon initial entry to the loop 81 * $1 is number of quadwords to clear (zero is a valid value) 82 * $2 is number of trailing bytes (0..7) ($2 never used...) 83 * $16 is known to be aligned 0mod8 84 */ 85 $headalign: 86 subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop 87 and $16, 0x3f, $2 # .. .. E .. : Forward work for huge loop 88 subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) 89 blt $4, $trailquad # U .. .. .. : U L U L 90 91 /* 92 * We know that we're going to do at least 16 quads, which means we are 93 * going to be able to use the large block clear loop at least once. 94 * Figure out how many quads we need to clear before we are 0mod64 aligned 95 * so we can use the wh64 instruction. 96 */ 97 98 nop # .. .. .. E 99 nop # .. .. E .. 100 nop # .. E .. .. 101 beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 102 103 $alignmod64: 104 EX( stq_u $31, 0($16) ) # .. .. .. L 105 addq $3, 8, $3 # .. .. E .. 106 subq $0, 8, $0 # .. E .. .. 107 nop # E .. .. .. : U L U L 108 109 nop # .. .. .. E 110 subq $1, 1, $1 # .. .. E .. 111 addq $16, 8, $16 # .. E .. .. 112 blt $3, $alignmod64 # U .. .. .. : U L U L 113 114 $bigalign: 115 /* 116 * $0 is the number of bytes left 117 * $1 is the number of quads left 118 * $16 is aligned 0mod64 119 * we know that we'll be taking a minimum of one trip through 120 * CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle 121 * We are _not_ going to update $0 after every single store. That 122 * would be silly, because there will be cross-cluster dependencies 123 * no matter how the code is scheduled. By doing it in slightly 124 * staggered fashion, we can still do this loop in 5 fetches 125 * The worse case will be doing two extra quads in some future execution, 126 * in the event of an interrupted clear. 127 * Assumes the wh64 needs to be for 2 trips through the loop in the future 128 * The wh64 is issued on for the starting destination address for trip +2 129 * through the loop, and if there are less than two trips left, the target 130 * address will be for the current trip. 131 */ 132 nop # E : 133 nop # E : 134 nop # E : 135 bis $16,$16,$3 # E : U L U L : Initial wh64 address is dest 136 /* This might actually help for the current trip... */ 137 138 $do_wh64: 139 wh64 ($3) # .. .. .. L1 : memory subsystem hint 140 subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? 141 EX( stq_u $31, 0($16) ) # .. L .. .. 142 subq $0, 8, $0 # E .. .. .. : U L U L 143 144 addq $16, 128, $3 # E : Target address of wh64 145 EX( stq_u $31, 8($16) ) # L : 146 EX( stq_u $31, 16($16) ) # L : 147 subq $0, 16, $0 # E : U L L U 148 149 nop # E : 150 EX( stq_u $31, 24($16) ) # L : 151 EX( stq_u $31, 32($16) ) # L : 152 subq $0, 168, $5 # E : U L L U : two trips through the loop left? 153 /* 168 = 192 - 24, since we've already completed some stores */ 154 155 subq $0, 16, $0 # E : 156 EX( stq_u $31, 40($16) ) # L : 157 EX( stq_u $31, 48($16) ) # L : 158 cmovlt $5, $16, $3 # E : U L L U : Latency 2, extra mapping cycle 159 160 subq $1, 8, $1 # E : 161 subq $0, 16, $0 # E : 162 EX( stq_u $31, 56($16) ) # L : 163 nop # E : U L U L 164 165 nop # E : 166 subq $0, 8, $0 # E : 167 addq $16, 64, $16 # E : 168 bge $4, $do_wh64 # U : U L U L 169 170 $trailquad: 171 # zero to 16 quadwords left to store, plus any trailing bytes 172 # $1 is the number of quadwords left to go. 173 # 174 nop # .. .. .. E 175 nop # .. .. E .. 176 nop # .. E .. .. 177 beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go 178 179 $onequad: 180 EX( stq_u $31, 0($16) ) # .. .. .. L 181 subq $1, 1, $1 # .. .. E .. 182 subq $0, 8, $0 # .. E .. .. 183 nop # E .. .. .. : U L U L 184 185 nop # .. .. .. E 186 nop # .. .. E .. 187 addq $16, 8, $16 # .. E .. .. 188 bgt $1, $onequad # U .. .. .. : U L U L 189 190 # We have an unknown number of bytes left to go. 191 $trailbytes: 192 nop # .. .. .. E 193 nop # .. .. E .. 194 nop # .. E .. .. 195 beq $0, $zerolength # U .. .. .. : U L U L 196 197 # $0 contains the number of bytes left to copy (0..31) 198 # so we will use $0 as the loop counter 199 # We know for a fact that $0 > 0 zero due to previous context 200 $onebyte: 201 EX( stb $31, 0($16) ) # .. .. .. L 202 subq $0, 1, $0 # .. .. E .. : 203 addq $16, 1, $16 # .. E .. .. : 204 bgt $0, $onebyte # U .. .. .. : U L U L 205 206 $zerolength: 207 $exception: # Destination for exception recovery(?) 208 nop # .. .. .. E : 209 nop # .. .. E .. : 210 nop # .. E .. .. : 211 ret $31, ($26), 1 # L0 .. .. .. : L U L U 212 .end __clear_user 213 EXPORT_SYMBOL(__clear_user) 214