xref: /openbmc/linux/arch/alpha/lib/ev6-memchr.S (revision 293d5b43)
1/*
2 * arch/alpha/lib/ev6-memchr.S
3 *
4 * 21264 version contributed by Rick Gorton <rick.gorton@alpha-processor.com>
5 *
6 * Finds characters in a memory area.  Optimized for the Alpha:
7 *
8 *    - memory accessed as aligned quadwords only
9 *    - uses cmpbge to compare 8 bytes in parallel
10 *    - does binary search to find 0 byte in last
11 *      quadword (HAKMEM needed 12 instructions to
12 *      do this instead of the 9 instructions that
13 *      binary search needs).
14 *
15 * For correctness consider that:
16 *
17 *    - only minimum number of quadwords may be accessed
18 *    - the third argument is an unsigned long
19 *
20 * Much of the information about 21264 scheduling/coding comes from:
21 *	Compiler Writer's Guide for the Alpha 21264
22 *	abbreviated as 'CWG' in other comments here
23 *	ftp.digital.com/pub/Digital/info/semiconductor/literature/dsc-library.html
24 * Scheduling notation:
25 *	E	- either cluster
26 *	U	- upper subcluster; U0 - subcluster U0; U1 - subcluster U1
27 *	L	- lower subcluster; L0 - subcluster L0; L1 - subcluster L1
28 * Try not to change the actual algorithm if possible for consistency.
29 */
30
31        .set noreorder
32        .set noat
33
34	.align	4
35	.globl memchr
36	.ent memchr
37memchr:
38	.frame $30,0,$26,0
39	.prologue 0
40
41	# Hack -- if someone passes in (size_t)-1, hoping to just
42	# search til the end of the address space, we will overflow
43	# below when we find the address of the last byte.  Given
44	# that we will never have a 56-bit address space, cropping
45	# the length is the easiest way to avoid trouble.
46	zap	$18, 0x80, $5	# U : Bound length
47	beq	$18, $not_found	# U :
48        ldq_u   $1, 0($16)	# L : load first quadword Latency=3
49	and	$17, 0xff, $17	# E : L L U U : 00000000000000ch
50
51	insbl	$17, 1, $2	# U : 000000000000ch00
52	cmpult	$18, 9, $4	# E : small (< 1 quad) string?
53	or	$2, $17, $17	# E : 000000000000chch
54        lda     $3, -1($31)	# E : U L L U
55
56	sll	$17, 16, $2	# U : 00000000chch0000
57	addq	$16, $5, $5	# E : Max search address
58	or	$2, $17, $17	# E : 00000000chchchch
59	sll	$17, 32, $2	# U : U L L U : chchchch00000000
60
61	or	$2, $17, $17	# E : chchchchchchchch
62	extql	$1, $16, $7	# U : $7 is upper bits
63	beq	$4, $first_quad	# U :
64	ldq_u	$6, -1($5)	# L : L U U L : eight or less bytes to search Latency=3
65
66	extqh	$6, $16, $6	# U : 2 cycle stall for $6
67	mov	$16, $0		# E :
68	nop			# E :
69	or	$7, $6, $1	# E : L U L U $1 = quadword starting at $16
70
71	# Deal with the case where at most 8 bytes remain to be searched
72	# in $1.  E.g.:
73	#	$18 = 6
74	#	$1 = ????c6c5c4c3c2c1
75$last_quad:
76	negq	$18, $6		# E :
77        xor	$17, $1, $1	# E :
78	srl	$3, $6, $6	# U : $6 = mask of $18 bits set
79        cmpbge  $31, $1, $2	# E : L U L U
80
81	nop
82	nop
83	and	$2, $6, $2	# E :
84        beq     $2, $not_found	# U : U L U L
85
86$found_it:
87#ifdef CONFIG_ALPHA_EV67
88	/*
89	 * Since we are guaranteed to have set one of the bits, we don't
90	 * have to worry about coming back with a 0x40 out of cttz...
91	 */
92	cttz	$2, $3		# U0 :
93	addq	$0, $3, $0	# E : All done
94	nop			# E :
95	ret			# L0 : L U L U
96#else
97	/*
98	 * Slow and clunky.  It can probably be improved.
99	 * An exercise left for others.
100	 */
101        negq    $2, $3		# E :
102        and     $2, $3, $2	# E :
103        and     $2, 0x0f, $1	# E :
104        addq    $0, 4, $3	# E :
105
106        cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
107	nop			# E : keep with cmov
108        and     $2, 0x33, $1	# E :
109        addq    $0, 2, $3	# E : U L U L : 2 cycle stall on $0
110
111        cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
112	nop			# E : keep with cmov
113        and     $2, 0x55, $1	# E :
114        addq    $0, 1, $3	# E : U L U L : 2 cycle stall on $0
115
116        cmoveq  $1, $3, $0	# E : Latency 2, extra map cycle
117	nop
118	nop
119	ret			# L0 : L U L U
120#endif
121
122	# Deal with the case where $18 > 8 bytes remain to be
123	# searched.  $16 may not be aligned.
124	.align 4
125$first_quad:
126	andnot	$16, 0x7, $0	# E :
127        insqh   $3, $16, $2	# U : $2 = 0000ffffffffffff ($16<0:2> ff)
128        xor	$1, $17, $1	# E :
129	or	$1, $2, $1	# E : U L U L $1 = ====ffffffffffff
130
131        cmpbge  $31, $1, $2	# E :
132        bne     $2, $found_it	# U :
133	# At least one byte left to process.
134	ldq	$1, 8($0)	# L :
135	subq	$5, 1, $18	# E : U L U L
136
137	addq	$0, 8, $0	# E :
138	# Make $18 point to last quad to be accessed (the
139	# last quad may or may not be partial).
140	andnot	$18, 0x7, $18	# E :
141	cmpult	$0, $18, $2	# E :
142	beq	$2, $final	# U : U L U L
143
144	# At least two quads remain to be accessed.
145
146	subq	$18, $0, $4	# E : $4 <- nr quads to be processed
147	and	$4, 8, $4	# E : odd number of quads?
148	bne	$4, $odd_quad_count # U :
149	# At least three quads remain to be accessed
150	mov	$1, $4		# E : L U L U : move prefetched value to correct reg
151
152	.align	4
153$unrolled_loop:
154	ldq	$1, 8($0)	# L : prefetch $1
155	xor	$17, $4, $2	# E :
156	cmpbge	$31, $2, $2	# E :
157	bne	$2, $found_it	# U : U L U L
158
159	addq	$0, 8, $0	# E :
160	nop			# E :
161	nop			# E :
162	nop			# E :
163
164$odd_quad_count:
165	xor	$17, $1, $2	# E :
166	ldq	$4, 8($0)	# L : prefetch $4
167	cmpbge	$31, $2, $2	# E :
168	addq	$0, 8, $6	# E :
169
170	bne	$2, $found_it	# U :
171	cmpult	$6, $18, $6	# E :
172	addq	$0, 8, $0	# E :
173	nop			# E :
174
175	bne	$6, $unrolled_loop # U :
176	mov	$4, $1		# E : move prefetched value into $1
177	nop			# E :
178	nop			# E :
179
180$final:	subq	$5, $0, $18	# E : $18 <- number of bytes left to do
181	nop			# E :
182	nop			# E :
183	bne	$18, $last_quad	# U :
184
185$not_found:
186	mov	$31, $0		# E :
187	nop			# E :
188	nop			# E :
189	ret			# L0 :
190
191        .end memchr
192