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/arch/alpha/lib/ev6-copy_user.S

http://github.com/mirrors/linux
Assembly | 227 lines | 210 code | 17 blank | 0 comment | 9 complexity | bbdb21782ff540305b7fc949f8577589 MD5 | raw file
  1/* SPDX-License-Identifier: GPL-2.0 */
  2/*
  3 * arch/alpha/lib/ev6-copy_user.S
  4 *
  5 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
  6 *
  7 * Copy to/from user space, handling exceptions as we go..  This
  8 * isn't exactly pretty.
  9 *
 10 * This is essentially the same as "memcpy()", but with a few twists.
 11 * Notably, we have to make sure that $0 is always up-to-date and
 12 * contains the right "bytes left to copy" value (and that it is updated
 13 * only _after_ a successful copy). There is also some rather minor
 14 * exception setup stuff..
 15 *
 16 * Much of the information about 21264 scheduling/coding comes from:
 17 *	Compiler Writer's Guide for the Alpha 21264
 18 *	abbreviated as 'CWG' in other comments here
 19 *	ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
 20 * Scheduling notation:
 21 *	E	- either cluster
 22 *	U	- upper subcluster; U0 - subcluster U0; U1 - subcluster U1
 23 *	L	- lower subcluster; L0 - subcluster L0; L1 - subcluster L1
 24 */
 25
 26#include <asm/export.h>
 27/* Allow an exception for an insn; exit if we get one.  */
 28#define EXI(x,y...)			\
 29	99: x,##y;			\
 30	.section __ex_table,"a";	\
 31	.long 99b - .;			\
 32	lda $31, $exitin-99b($31);	\
 33	.previous
 34
 35#define EXO(x,y...)			\
 36	99: x,##y;			\
 37	.section __ex_table,"a";	\
 38	.long 99b - .;			\
 39	lda $31, $exitout-99b($31);	\
 40	.previous
 41
 42	.set noat
 43	.align 4
 44	.globl __copy_user
 45	.ent __copy_user
 46				# Pipeline info: Slotting & Comments
 47__copy_user:
 48	.prologue 0
 49	mov $18, $0		# .. .. .. E
 50	subq $18, 32, $1	# .. .. E. ..	: Is this going to be a small copy?
 51	nop			# .. E  .. ..
 52	beq $18, $zerolength	# U  .. .. ..	: U L U L
 53
 54	and $16,7,$3		# .. .. .. E	: is leading dest misalignment
 55	ble $1, $onebyteloop	# .. .. U  ..	: 1st branch : small amount of data
 56	beq $3, $destaligned	# .. U  .. ..	: 2nd (one cycle fetcher stall)
 57	subq $3, 8, $3		# E  .. .. ..	: L U U L : trip counter
 58/*
 59 * The fetcher stall also hides the 1 cycle cross-cluster stall for $3 (L --> U)
 60 * This loop aligns the destination a byte at a time
 61 * We know we have at least one trip through this loop
 62 */
 63$aligndest:
 64	EXI( ldbu $1,0($17) )	# .. .. .. L	: Keep loads separate from stores
 65	addq $16,1,$16		# .. .. E  ..	: Section 3.8 in the CWG
 66	addq $3,1,$3		# .. E  .. ..	:
 67	nop			# E  .. .. ..	: U L U L
 68
 69/*
 70 * the -1 is to compensate for the inc($16) done in a previous quadpack
 71 * which allows us zero dependencies within either quadpack in the loop
 72 */
 73	EXO( stb $1,-1($16) )	# .. .. .. L	:
 74	addq $17,1,$17		# .. .. E  ..	: Section 3.8 in the CWG
 75	subq $0,1,$0		# .. E  .. ..	:
 76	bne $3, $aligndest	# U  .. .. ..	: U L U L
 77
 78/*
 79 * If we fell through into here, we have a minimum of 33 - 7 bytes
 80 * If we arrived via branch, we have a minimum of 32 bytes
 81 */
 82$destaligned:
 83	and $17,7,$1		# .. .. .. E	: Check _current_ source alignment
 84	bic $0,7,$4		# .. .. E  ..	: number bytes as a quadword loop
 85	EXI( ldq_u $3,0($17) )	# .. L  .. ..	: Forward fetch for fallthrough code
 86	beq $1,$quadaligned	# U  .. .. ..	: U L U L
 87
 88/*
 89 * In the worst case, we've just executed an ldq_u here from 0($17)
 90 * and we'll repeat it once if we take the branch
 91 */
 92
 93/* Misaligned quadword loop - not unrolled.  Leave it that way. */
 94$misquad:
 95	EXI( ldq_u $2,8($17) )	# .. .. .. L	:
 96	subq $4,8,$4		# .. .. E  ..	:
 97	extql $3,$17,$3		# .. U  .. ..	:
 98	extqh $2,$17,$1		# U  .. .. ..	: U U L L
 99
100	bis $3,$1,$1		# .. .. .. E	:
101	EXO( stq $1,0($16) )	# .. .. L  ..	:
102	addq $17,8,$17		# .. E  .. ..	:
103	subq $0,8,$0		# E  .. .. ..	: U L L U
104
105	addq $16,8,$16		# .. .. .. E	:
106	bis $2,$2,$3		# .. .. E  ..	:
107	nop			# .. E  .. ..	:
108	bne $4,$misquad		# U  .. .. ..	: U L U L
109
110	nop			# .. .. .. E
111	nop			# .. .. E  ..
112	nop			# .. E  .. ..
113	beq $0,$zerolength	# U  .. .. ..	: U L U L
114
115/* We know we have at least one trip through the byte loop */
116	EXI ( ldbu $2,0($17) )	# .. .. .. L	: No loads in the same quad
117	addq $16,1,$16		# .. .. E  ..	: as the store (Section 3.8 in CWG)
118	nop			# .. E  .. ..	:
119	br $31, $dirtyentry	# L0 .. .. ..	: L U U L
120/* Do the trailing byte loop load, then hop into the store part of the loop */
121
122/*
123 * A minimum of (33 - 7) bytes to do a quad at a time.
124 * Based upon the usage context, it's worth the effort to unroll this loop
125 * $0 - number of bytes to be moved
126 * $4 - number of bytes to move as quadwords
127 * $16 is current destination address
128 * $17 is current source address
129 */
130$quadaligned:
131	subq	$4, 32, $2	# .. .. .. E	: do not unroll for small stuff
132	nop			# .. .. E  ..
133	nop			# .. E  .. ..
134	blt	$2, $onequad	# U  .. .. ..	: U L U L
135
136/*
137 * There is a significant assumption here that the source and destination
138 * addresses differ by more than 32 bytes.  In this particular case, a
139 * sparsity of registers further bounds this to be a minimum of 8 bytes.
140 * But if this isn't met, then the output result will be incorrect.
141 * Furthermore, due to a lack of available registers, we really can't
142 * unroll this to be an 8x loop (which would enable us to use the wh64
143 * instruction memory hint instruction).
144 */
145$unroll4:
146	EXI( ldq $1,0($17) )	# .. .. .. L
147	EXI( ldq $2,8($17) )	# .. .. L  ..
148	subq	$4,32,$4	# .. E  .. ..
149	nop			# E  .. .. ..	: U U L L
150
151	addq	$17,16,$17	# .. .. .. E
152	EXO( stq $1,0($16) )	# .. .. L  ..
153	EXO( stq $2,8($16) )	# .. L  .. ..
154	subq	$0,16,$0	# E  .. .. ..	: U L L U
155
156	addq	$16,16,$16	# .. .. .. E
157	EXI( ldq $1,0($17) )	# .. .. L  ..
158	EXI( ldq $2,8($17) )	# .. L  .. ..
159	subq	$4, 32, $3	# E  .. .. ..	: U U L L : is there enough for another trip?
160
161	EXO( stq $1,0($16) )	# .. .. .. L
162	EXO( stq $2,8($16) )	# .. .. L  ..
163	subq	$0,16,$0	# .. E  .. ..
164	addq	$17,16,$17	# E  .. .. ..	: U L L U
165
166	nop			# .. .. .. E
167	nop			# .. .. E  ..
168	addq	$16,16,$16	# .. E  .. ..
169	bgt	$3,$unroll4	# U  .. .. ..	: U L U L
170
171	nop
172	nop
173	nop
174	beq	$4, $noquads
175
176$onequad:
177	EXI( ldq $1,0($17) )
178	subq	$4,8,$4
179	addq	$17,8,$17
180	nop
181
182	EXO( stq $1,0($16) )
183	subq	$0,8,$0
184	addq	$16,8,$16
185	bne	$4,$onequad
186
187$noquads:
188	nop
189	nop
190	nop
191	beq $0,$zerolength
192
193/*
194 * For small copies (or the tail of a larger copy), do a very simple byte loop.
195 * There's no point in doing a lot of complex alignment calculations to try to
196 * to quadword stuff for a small amount of data.
197 *	$0 - remaining number of bytes left to copy
198 *	$16 - current dest addr
199 *	$17 - current source addr
200 */
201
202$onebyteloop:
203	EXI ( ldbu $2,0($17) )	# .. .. .. L	: No loads in the same quad
204	addq $16,1,$16		# .. .. E  ..	: as the store (Section 3.8 in CWG)
205	nop			# .. E  .. ..	:
206	nop			# E  .. .. ..	: U L U L
207
208$dirtyentry:
209/*
210 * the -1 is to compensate for the inc($16) done in a previous quadpack
211 * which allows us zero dependencies within either quadpack in the loop
212 */
213	EXO ( stb $2,-1($16) )	# .. .. .. L	:
214	addq $17,1,$17		# .. .. E  ..	: quadpack as the load
215	subq $0,1,$0		# .. E  .. ..	: change count _after_ copy
216	bgt $0,$onebyteloop	# U  .. .. ..	: U L U L
217
218$zerolength:
219$exitin:
220$exitout:			# Destination for exception recovery(?)
221	nop			# .. .. .. E
222	nop			# .. .. E  ..
223	nop			# .. E  .. ..
224	ret $31,($26),1		# L0 .. .. ..	: L U L U
225
226	.end __copy_user
227	EXPORT_SYMBOL(__copy_user)