214 lines
7 KiB
ArmAsm
214 lines
7 KiB
ArmAsm
/* 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 <asm/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)
|