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213 lines
7.0 KiB
213 lines
7.0 KiB
/* SPDX-License-Identifier: GPL-2.0 */ |
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/* |
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* arch/alpha/lib/ev6-clear_user.S |
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* 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com> |
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* |
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* Zero user space, handling exceptions as we go. |
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* |
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* We have to make sure that $0 is always up-to-date and contains the |
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* right "bytes left to zero" value (and that it is updated only _after_ |
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* a successful copy). There is also some rather minor exception setup |
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* stuff. |
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* |
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* Much of the information about 21264 scheduling/coding comes from: |
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* Compiler Writer's Guide for the Alpha 21264 |
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* abbreviated as 'CWG' in other comments here |
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* ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html |
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* Scheduling notation: |
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* E - either cluster |
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* U - upper subcluster; U0 - subcluster U0; U1 - subcluster U1 |
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* L - lower subcluster; L0 - subcluster L0; L1 - subcluster L1 |
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* Try not to change the actual algorithm if possible for consistency. |
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* Determining actual stalls (other than slotting) doesn't appear to be easy to do. |
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* From perusing the source code context where this routine is called, it is |
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* a fair assumption that significant fractions of entire pages are zeroed, so |
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* it's going to be worth the effort to hand-unroll a big loop, and use wh64. |
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* ASSUMPTION: |
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* The believed purpose of only updating $0 after a store is that a signal |
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* may come along during the execution of this chunk of code, and we don't |
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* want to leave a hole (and we also want to avoid repeating lots of work) |
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*/ |
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#include <asm/export.h> |
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/* Allow an exception for an insn; exit if we get one. */ |
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#define EX(x,y...) \ |
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99: x,##y; \ |
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.section __ex_table,"a"; \ |
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.long 99b - .; \ |
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lda $31, $exception-99b($31); \ |
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.previous |
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.set noat |
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.set noreorder |
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.align 4 |
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.globl __clear_user |
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.ent __clear_user |
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.frame $30, 0, $26 |
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.prologue 0 |
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# Pipeline info : Slotting & Comments |
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__clear_user: |
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and $17, $17, $0 |
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and $16, 7, $4 # .. E .. .. : find dest head misalignment |
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beq $0, $zerolength # U .. .. .. : U L U L |
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addq $0, $4, $1 # .. .. .. E : bias counter |
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and $1, 7, $2 # .. .. E .. : number of misaligned bytes in tail |
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# Note - we never actually use $2, so this is a moot computation |
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# and we can rewrite this later... |
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srl $1, 3, $1 # .. E .. .. : number of quadwords to clear |
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beq $4, $headalign # U .. .. .. : U L U L |
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/* |
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* Head is not aligned. Write (8 - $4) bytes to head of destination |
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* This means $16 is known to be misaligned |
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*/ |
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EX( ldq_u $5, 0($16) ) # .. .. .. L : load dst word to mask back in |
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beq $1, $onebyte # .. .. U .. : sub-word store? |
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mskql $5, $16, $5 # .. U .. .. : take care of misaligned head |
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addq $16, 8, $16 # E .. .. .. : L U U L |
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EX( stq_u $5, -8($16) ) # .. .. .. L : |
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subq $1, 1, $1 # .. .. E .. : |
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addq $0, $4, $0 # .. E .. .. : bytes left -= 8 - misalignment |
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subq $0, 8, $0 # E .. .. .. : U L U L |
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.align 4 |
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/* |
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* (The .align directive ought to be a moot point) |
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* values upon initial entry to the loop |
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* $1 is number of quadwords to clear (zero is a valid value) |
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* $2 is number of trailing bytes (0..7) ($2 never used...) |
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* $16 is known to be aligned 0mod8 |
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*/ |
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$headalign: |
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subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop |
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and $16, 0x3f, $2 # .. .. E .. : Forward work for huge loop |
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subq $2, 0x40, $3 # .. E .. .. : bias counter (huge loop) |
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blt $4, $trailquad # U .. .. .. : U L U L |
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/* |
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* We know that we're going to do at least 16 quads, which means we are |
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* going to be able to use the large block clear loop at least once. |
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* Figure out how many quads we need to clear before we are 0mod64 aligned |
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* so we can use the wh64 instruction. |
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*/ |
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nop # .. .. .. E |
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nop # .. .. E .. |
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nop # .. E .. .. |
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beq $3, $bigalign # U .. .. .. : U L U L : Aligned 0mod64 |
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$alignmod64: |
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EX( stq_u $31, 0($16) ) # .. .. .. L |
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addq $3, 8, $3 # .. .. E .. |
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subq $0, 8, $0 # .. E .. .. |
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nop # E .. .. .. : U L U L |
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nop # .. .. .. E |
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subq $1, 1, $1 # .. .. E .. |
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addq $16, 8, $16 # .. E .. .. |
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blt $3, $alignmod64 # U .. .. .. : U L U L |
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$bigalign: |
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/* |
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* $0 is the number of bytes left |
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* $1 is the number of quads left |
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* $16 is aligned 0mod64 |
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* we know that we'll be taking a minimum of one trip through |
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* CWG Section 3.7.6: do not expect a sustained store rate of > 1/cycle |
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* We are _not_ going to update $0 after every single store. That |
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* would be silly, because there will be cross-cluster dependencies |
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* no matter how the code is scheduled. By doing it in slightly |
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* staggered fashion, we can still do this loop in 5 fetches |
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* The worse case will be doing two extra quads in some future execution, |
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* in the event of an interrupted clear. |
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* Assumes the wh64 needs to be for 2 trips through the loop in the future |
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* The wh64 is issued on for the starting destination address for trip +2 |
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* through the loop, and if there are less than two trips left, the target |
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* address will be for the current trip. |
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*/ |
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nop # E : |
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nop # E : |
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nop # E : |
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bis $16,$16,$3 # E : U L U L : Initial wh64 address is dest |
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/* This might actually help for the current trip... */ |
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$do_wh64: |
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wh64 ($3) # .. .. .. L1 : memory subsystem hint |
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subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? |
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EX( stq_u $31, 0($16) ) # .. L .. .. |
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subq $0, 8, $0 # E .. .. .. : U L U L |
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addq $16, 128, $3 # E : Target address of wh64 |
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EX( stq_u $31, 8($16) ) # L : |
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EX( stq_u $31, 16($16) ) # L : |
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subq $0, 16, $0 # E : U L L U |
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nop # E : |
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EX( stq_u $31, 24($16) ) # L : |
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EX( stq_u $31, 32($16) ) # L : |
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subq $0, 168, $5 # E : U L L U : two trips through the loop left? |
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/* 168 = 192 - 24, since we've already completed some stores */ |
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subq $0, 16, $0 # E : |
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EX( stq_u $31, 40($16) ) # L : |
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EX( stq_u $31, 48($16) ) # L : |
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cmovlt $5, $16, $3 # E : U L L U : Latency 2, extra mapping cycle |
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subq $1, 8, $1 # E : |
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subq $0, 16, $0 # E : |
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EX( stq_u $31, 56($16) ) # L : |
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nop # E : U L U L |
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nop # E : |
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subq $0, 8, $0 # E : |
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addq $16, 64, $16 # E : |
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bge $4, $do_wh64 # U : U L U L |
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$trailquad: |
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# zero to 16 quadwords left to store, plus any trailing bytes |
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# $1 is the number of quadwords left to go. |
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# |
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nop # .. .. .. E |
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nop # .. .. E .. |
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nop # .. E .. .. |
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beq $1, $trailbytes # U .. .. .. : U L U L : Only 0..7 bytes to go |
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$onequad: |
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EX( stq_u $31, 0($16) ) # .. .. .. L |
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subq $1, 1, $1 # .. .. E .. |
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subq $0, 8, $0 # .. E .. .. |
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nop # E .. .. .. : U L U L |
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nop # .. .. .. E |
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nop # .. .. E .. |
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addq $16, 8, $16 # .. E .. .. |
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bgt $1, $onequad # U .. .. .. : U L U L |
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# We have an unknown number of bytes left to go. |
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$trailbytes: |
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nop # .. .. .. E |
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nop # .. .. E .. |
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nop # .. E .. .. |
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beq $0, $zerolength # U .. .. .. : U L U L |
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# $0 contains the number of bytes left to copy (0..31) |
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# so we will use $0 as the loop counter |
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# We know for a fact that $0 > 0 zero due to previous context |
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$onebyte: |
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EX( stb $31, 0($16) ) # .. .. .. L |
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subq $0, 1, $0 # .. .. E .. : |
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addq $16, 1, $16 # .. E .. .. : |
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bgt $0, $onebyte # U .. .. .. : U L U L |
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$zerolength: |
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$exception: # Destination for exception recovery(?) |
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nop # .. .. .. E : |
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nop # .. .. E .. : |
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nop # .. E .. .. : |
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ret $31, ($26), 1 # L0 .. .. .. : L U L U |
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.end __clear_user |
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EXPORT_SYMBOL(__clear_user)
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