arch/alpha/lib/ev6-copy_user.S - linux - Git at Google

 /* SPDX-License-Identifier: GPL-2.0 */
 /*
  * arch/alpha/lib/ev6-copy_user.S
  *
  * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
  *
  * 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)
	/* SPDX-License-Identifier: GPL-2.0 */
	/*
	* arch/alpha/lib/ev6-copy_user.S
	*
	* 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
	*
	* 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)