/* SPDX-License-Identifier: GPL-2.0 */
/*
* arch/alpha/lib/ev6-copy_user.S
*
* 21264 version contributed by Rick Gorton <[email protected]>
*
* Copy to/from user space, handling exceptions as we go.. This
* isn't exactly pretty.
*
* This is essentially the same as "memcpy()", but with a few twists.
* Notably, we have to make sure that $0 is always up-to-date and
* contains the right "bytes left to copy" 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
*/
#include <linux/export.h>
/* Allow an exception for an insn; exit if we get one. */
#define EXI(x,y...) \
99: x,##y; \
.section __ex_table,"a"; \
.long 99b - .; \
lda $31, $exitin-99b($31); \
.previous
#define EXO(x,y...) \
99: x,##y; \
.section __ex_table,"a"; \
.long 99b - .; \
lda $31, $exitout-99b($31); \
.previous
.set noat
.align 4
.globl __copy_user
.ent __copy_user
# Pipeline info: Slotting & Comments
__copy_user:
.prologue 0
mov $18, $0 # .. .. .. E
subq $18, 32, $1 # .. .. E. .. : Is this going to be a small copy?
nop # .. E .. ..
beq $18, $zerolength # U .. .. .. : U L U L
and $16,7,$3 # .. .. .. E : is leading dest misalignment
ble $1, $onebyteloop # .. .. U .. : 1st branch : small amount of data
beq $3, $destaligned # .. U .. .. : 2nd (one cycle fetcher stall)
subq $3, 8, $3 # E .. .. .. : L U U L : trip counter
/*
* The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U)
* This loop aligns the destination a byte at a time
* We know we have at least one trip through this loop
*/
$aligndest:
EXI( ldbu $1,0($17) ) # .. .. .. L : Keep loads separate from stores
addq $16,1,$16 # .. .. E .. : Section 3.8 in the CWG
addq $3,1,$3 # .. E .. .. :
nop # E .. .. .. : U L U L
/*
* the -1 is to compensate for the inc($16) done in a previous quadpack
* which allows us zero dependencies within either quadpack in the loop
*/
EXO( stb $1,-1($16) ) # .. .. .. L :
addq $17,1,$17 # .. .. E .. : Section 3.8 in the CWG
subq $0,1,$0 # .. E .. .. :
bne $3, $aligndest # U .. .. .. : U L U L
/*
* If we fell through into here, we have a minimum of 33 - 7 bytes
* If we arrived via branch, we have a minimum of 32 bytes
*/
$destaligned:
and $17,7,$1 # .. .. .. E : Check _current_ source alignment
bic $0,7,$4 # .. .. E .. : number bytes as a quadword loop
EXI( ldq_u $3,0($17) ) # .. L .. .. : Forward fetch for fallthrough code
beq $1,$quadaligned # U .. .. .. : U L U L
/*
* In the worst case, we've just executed an ldq_u here from 0($17)
* and we'll repeat it once if we take the branch
*/
/* Misaligned quadword loop - not unrolled. Leave it that way. */
$misquad:
EXI( ldq_u $2,8($17) ) # .. .. .. L :
subq $4,8,$4 # .. .. E .. :
extql $3,$17,$3 # .. U .. .. :
extqh $2,$17,$1 # U .. .. .. : U U L L
bis $3,$1,$1 # .. .. .. E :
EXO( stq $1,0($16) ) # .. .. L .. :
addq $17,8,$17 # .. E .. .. :
subq $0,8,$0 # E .. .. .. : U L L U
addq $16,8,$16 # .. .. .. E :
bis $2,$2,$3 # .. .. E .. :
nop # .. E .. .. :
bne $4,$misquad # U .. .. .. : U L U L
nop # .. .. .. E
nop # .. .. E ..
nop # .. E .. ..
beq $0,$zerolength # U .. .. .. : U L U L
/* We know we have at least one trip through the byte loop */
EXI ( ldbu $2,0($17) ) # .. .. .. L : No loads in the same quad
addq $16,1,$16 # .. .. E .. : as the store (Section 3.8 in CWG)
nop # .. E .. .. :
br $31, $dirtyentry # L0 .. .. .. : L U U L
/* Do the trailing byte loop load, then hop into the store part of the loop */
/*
* A minimum of (33 - 7) bytes to do a quad at a time.
* Based upon the usage context, it's worth the effort to unroll this loop
* $0 - number of bytes to be moved
* $4 - number of bytes to move as quadwords
* $16 is current destination address
* $17 is current source address
*/
$quadaligned:
subq $4, 32, $2 # .. .. .. E : do not unroll for small stuff
nop # .. .. E ..
nop # .. E .. ..
blt $2, $onequad # U .. .. .. : U L U L
/*
* There is a significant assumption here that the source and destination
* addresses differ by more than 32 bytes. In this particular case, a
* sparsity of registers further bounds this to be a minimum of 8 bytes.
* But if this isn't met, then the output result will be incorrect.
* Furthermore, due to a lack of available registers, we really can't
* unroll this to be an 8x loop (which would enable us to use the wh64
* instruction memory hint instruction).
*/
$unroll4:
EXI( ldq $1,0($17) ) # .. .. .. L
EXI( ldq $2,8($17) ) # .. .. L ..
subq $4,32,$4 # .. E .. ..
nop # E .. .. .. : U U L L
addq $17,16,$17 # .. .. .. E
EXO( stq $1,0($16) ) # .. .. L ..
EXO( stq $2,8($16) ) # .. L .. ..
subq $0,16,$0 # E .. .. .. : U L L U
addq $16,16,$16 # .. .. .. E
EXI( ldq $1,0($17) ) # .. .. L ..
EXI( ldq $2,8($17) ) # .. L .. ..
subq $4, 32, $3 # E .. .. .. : U U L L : is there enough for another trip?
EXO( stq $1,0($16) ) # .. .. .. L
EXO( stq $2,8($16) ) # .. .. L ..
subq $0,16,$0 # .. E .. ..
addq $17,16,$17 # E .. .. .. : U L L U
nop # .. .. .. E
nop # .. .. E ..
addq $16,16,$16 # .. E .. ..
bgt $3,$unroll4 # U .. .. .. : U L U L
nop
nop
nop
beq $4, $noquads
$onequad:
EXI( ldq $1,0($17) )
subq $4,8,$4
addq $17,8,$17
nop
EXO( stq $1,0($16) )
subq $0,8,$0
addq $16,8,$16
bne $4,$onequad
$noquads:
nop
nop
nop
beq $0,$zerolength
/*
* For small copies (or the tail of a larger copy), do a very simple byte loop.
* There's no point in doing a lot of complex alignment calculations to try to
* to quadword stuff for a small amount of data.
* $0 - remaining number of bytes left to copy
* $16 - current dest addr
* $17 - current source addr
*/
$onebyteloop:
EXI ( ldbu $2,0($17) ) # .. .. .. L : No loads in the same quad
addq $16,1,$16 # .. .. E .. : as the store (Section 3.8 in CWG)
nop # .. E .. .. :
nop # E .. .. .. : U L U L
$dirtyentry:
/*
* the -1 is to compensate for the inc($16) done in a previous quadpack
* which allows us zero dependencies within either quadpack in the loop
*/
EXO ( stb $2,-1($16) ) # .. .. .. L :
addq $17,1,$17 # .. .. E .. : quadpack as the load
subq $0,1,$0 # .. E .. .. : change count _after_ copy
bgt $0,$onebyteloop # U .. .. .. : U L U L
$zerolength:
$exitin:
$exitout: # Destination for exception recovery(?)
nop # .. .. .. E
nop # .. .. E ..
nop # .. E .. ..
ret $31,($26),1 # L0 .. .. .. : L U L U
.end __copy_user
EXPORT_SYMBOL(__copy_user)
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