diff options
Diffstat (limited to 'ANDROID_3.4.5/arch/alpha/lib/ev6-clear_user.S')
| -rw-r--r-- | ANDROID_3.4.5/arch/alpha/lib/ev6-clear_user.S | 225 |
1 files changed, 0 insertions, 225 deletions
diff --git a/ANDROID_3.4.5/arch/alpha/lib/ev6-clear_user.S b/ANDROID_3.4.5/arch/alpha/lib/ev6-clear_user.S deleted file mode 100644 index 4f42a16b..00000000 --- a/ANDROID_3.4.5/arch/alpha/lib/ev6-clear_user.S +++ /dev/null @@ -1,225 +0,0 @@ -/* - * 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. - * - * NOTE! This is not directly C-callable, because the calling semantics - * are different: - * - * Inputs: - * length in $0 - * destination address in $6 - * exception pointer in $7 - * return address in $28 (exceptions expect it there) - * - * Outputs: - * bytes left to copy in $0 - * - * Clobbers: - * $1,$2,$3,$4,$5,$6 - * - * 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) - */ - -/* 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 __do_clear_user - .ent __do_clear_user - .frame $30, 0, $28 - .prologue 0 - - # Pipeline info : Slotting & Comments -__do_clear_user: - and $6, 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 $6 is known to be misaligned - */ - EX( ldq_u $5, 0($6) ) # .. .. .. L : load dst word to mask back in - beq $1, $onebyte # .. .. U .. : sub-word store? - mskql $5, $6, $5 # .. U .. .. : take care of misaligned head - addq $6, 8, $6 # E .. .. .. : L U U L - - EX( stq_u $5, -8($6) ) # .. .. .. 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...) - * $6 is known to be aligned 0mod8 - */ -$headalign: - subq $1, 16, $4 # .. .. .. E : If < 16, we can not use the huge loop - and $6, 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($6) ) # .. .. .. 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 $6, 8, $6 # .. E .. .. - blt $3, $alignmod64 # U .. .. .. : U L U L - -$bigalign: -/* - * $0 is the number of bytes left - * $1 is the number of quads left - * $6 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 $6,$6,$3 # E : U L U L : Initial wh64 address is dest - /* This might actually help for the current trip... */ - -$do_wh64: - wh64 ($3) # .. .. .. L1 : memory subsystem hint - subq $1, 16, $4 # .. .. E .. : Forward calculation - repeat the loop? - EX( stq_u $31, 0($6) ) # .. L .. .. - subq $0, 8, $0 # E .. .. .. : U L U L - - addq $6, 128, $3 # E : Target address of wh64 - EX( stq_u $31, 8($6) ) # L : - EX( stq_u $31, 16($6) ) # L : - subq $0, 16, $0 # E : U L L U - - nop # E : - EX( stq_u $31, 24($6) ) # L : - EX( stq_u $31, 32($6) ) # 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($6) ) # L : - EX( stq_u $31, 48($6) ) # L : - cmovlt $5, $6, $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($6) ) # L : - nop # E : U L U L - - nop # E : - subq $0, 8, $0 # E : - addq $6, 64, $6 # 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($6) ) # .. .. .. L - subq $1, 1, $1 # .. .. E .. - subq $0, 8, $0 # .. E .. .. - nop # E .. .. .. : U L U L - - nop # .. .. .. E - nop # .. .. E .. - addq $6, 8, $6 # .. 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($6) ) # .. .. .. L - subq $0, 1, $0 # .. .. E .. : - addq $6, 1, $6 # .. E .. .. : - bgt $0, $onebyte # U .. .. .. : U L U L - -$zerolength: -$exception: # Destination for exception recovery(?) - nop # .. .. .. E : - nop # .. .. E .. : - nop # .. E .. .. : - ret $31, ($28), 1 # L0 .. .. .. : L U L U - .end __do_clear_user - |