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