/*
* 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
