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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#if defined(__alpha_fix__) && defined(__alpha_cix__)
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
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