1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
|
/* Optimized strlen implementation for PowerPC.
Copyright (C) 1997-2020 Free Software Foundation, Inc.
This file is part of the GNU C Library.
The GNU C Library is free software; you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public
License as published by the Free Software Foundation; either
version 2.1 of the License, or (at your option) any later version.
The GNU C Library is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
Lesser General Public License for more details.
You should have received a copy of the GNU Lesser General Public
License along with the GNU C Library; if not, see
<https://www.gnu.org/licenses/>. */
#include <sysdep.h>
/* The algorithm here uses the following techniques:
1) Given a word 'x', we can test to see if it contains any 0 bytes
by subtracting 0x01010101, and seeing if any of the high bits of each
byte changed from 0 to 1. This works because the least significant
0 byte must have had no incoming carry (otherwise it's not the least
significant), so it is 0x00 - 0x01 == 0xff. For all other
byte values, either they have the high bit set initially, or when
1 is subtracted you get a value in the range 0x00-0x7f, none of which
have their high bit set. The expression here is
(x + 0xfefefeff) & ~(x | 0x7f7f7f7f), which gives 0x00000000 when
there were no 0x00 bytes in the word. You get 0x80 in bytes that
match, but possibly false 0x80 matches in the next more significant
byte to a true match due to carries. For little-endian this is
of no consequence since the least significant match is the one
we're interested in, but big-endian needs method 2 to find which
byte matches.
2) Given a word 'x', we can test to see _which_ byte was zero by
calculating ~(((x & 0x7f7f7f7f) + 0x7f7f7f7f) | x | 0x7f7f7f7f).
This produces 0x80 in each byte that was zero, and 0x00 in all
the other bytes. The '| 0x7f7f7f7f' clears the low 7 bits in each
byte, and the '| x' part ensures that bytes with the high bit set
produce 0x00. The addition will carry into the high bit of each byte
iff that byte had one of its low 7 bits set. We can then just see
which was the most significant bit set and divide by 8 to find how
many to add to the index.
This is from the book 'The PowerPC Compiler Writer's Guide',
by Steve Hoxey, Faraydon Karim, Bill Hay and Hank Warren.
We deal with strings not aligned to a word boundary by taking the
first word and ensuring that bytes not part of the string
are treated as nonzero. To allow for memory latency, we unroll the
loop a few times, being careful to ensure that we do not read ahead
across cache line boundaries.
Questions to answer:
1) How long are strings passed to strlen? If they're often really long,
we should probably use cache management instructions and/or unroll the
loop more. If they're often quite short, it might be better to use
fact (2) in the inner loop than have to recalculate it.
2) How popular are bytes with the high bit set? If they are very rare,
on some processors it might be useful to use the simpler expression
~((x - 0x01010101) | 0x7f7f7f7f) (that is, on processors with only one
ALU), but this fails when any character has its high bit set. */
/* Some notes on register usage: Under the SVR4 ABI, we can use registers
0 and 3 through 12 (so long as we don't call any procedures) without
saving them. We can also use registers 14 through 31 if we save them.
We can't use r1 (it's the stack pointer), r2 nor r13 because the user
program may expect them to hold their usual value if we get sent
a signal. Integer parameters are passed in r3 through r10.
We can use condition registers cr0, cr1, cr5, cr6, and cr7 without saving
them, the others we must save. */
/* int [r3] strlen (char *s [r3]) */
ENTRY (strlen)
#define rTMP4 r0
#define rRTN r3 /* incoming STR arg, outgoing result */
#define rSTR r4 /* current string position */
#define rPADN r5 /* number of padding bits we prepend to the
string to make it start at a word boundary */
#define rFEFE r6 /* constant 0xfefefeff (-0x01010101) */
#define r7F7F r7 /* constant 0x7f7f7f7f */
#define rWORD1 r8 /* current string word */
#define rWORD2 r9 /* next string word */
#define rMASK r9 /* mask for first string word */
#define rTMP1 r10
#define rTMP2 r11
#define rTMP3 r12
clrrwi rSTR, rRTN, 2
lis r7F7F, 0x7f7f
rlwinm rPADN, rRTN, 3, 27, 28
lwz rWORD1, 0(rSTR)
li rMASK, -1
addi r7F7F, r7F7F, 0x7f7f
/* We use method (2) on the first two words, because rFEFE isn't
required which reduces setup overhead. Also gives a faster return
for small strings on big-endian due to needing to recalculate with
method (2) anyway. */
#ifdef __LITTLE_ENDIAN__
slw rMASK, rMASK, rPADN
#else
srw rMASK, rMASK, rPADN
#endif
and rTMP1, r7F7F, rWORD1
or rTMP2, r7F7F, rWORD1
add rTMP1, rTMP1, r7F7F
nor rTMP3, rTMP2, rTMP1
and. rTMP3, rTMP3, rMASK
mtcrf 0x01, rRTN
bne L(done0)
lis rFEFE, -0x101
addi rFEFE, rFEFE, -0x101
/* Are we now aligned to a doubleword boundary? */
bt 29, L(loop)
/* Handle second word of pair. */
/* Perhaps use method (1) here for little-endian, saving one instruction? */
lwzu rWORD1, 4(rSTR)
and rTMP1, r7F7F, rWORD1
or rTMP2, r7F7F, rWORD1
add rTMP1, rTMP1, r7F7F
nor. rTMP3, rTMP2, rTMP1
bne L(done0)
/* The loop. */
L(loop):
lwz rWORD1, 4(rSTR)
lwzu rWORD2, 8(rSTR)
add rTMP1, rFEFE, rWORD1
nor rTMP2, r7F7F, rWORD1
and. rTMP1, rTMP1, rTMP2
add rTMP3, rFEFE, rWORD2
nor rTMP4, r7F7F, rWORD2
bne L(done1)
and. rTMP3, rTMP3, rTMP4
beq L(loop)
#ifndef __LITTLE_ENDIAN__
and rTMP1, r7F7F, rWORD2
add rTMP1, rTMP1, r7F7F
andc rTMP3, rTMP4, rTMP1
b L(done0)
L(done1):
and rTMP1, r7F7F, rWORD1
subi rSTR, rSTR, 4
add rTMP1, rTMP1, r7F7F
andc rTMP3, rTMP2, rTMP1
/* When we get to here, rSTR points to the first word in the string that
contains a zero byte, and rTMP3 has 0x80 for bytes that are zero,
and 0x00 otherwise. */
L(done0):
cntlzw rTMP3, rTMP3
subf rTMP1, rRTN, rSTR
srwi rTMP3, rTMP3, 3
add rRTN, rTMP1, rTMP3
blr
#else
L(done0):
addi rTMP1, rTMP3, -1 /* Form a mask from trailing zeros. */
andc rTMP1, rTMP1, rTMP3
cntlzw rTMP1, rTMP1 /* Count bits not in the mask. */
subf rTMP3, rRTN, rSTR
subfic rTMP1, rTMP1, 32-7
srwi rTMP1, rTMP1, 3
add rRTN, rTMP1, rTMP3
blr
L(done1):
addi rTMP3, rTMP1, -1
andc rTMP3, rTMP3, rTMP1
cntlzw rTMP3, rTMP3
subf rTMP1, rRTN, rSTR
subfic rTMP3, rTMP3, 32-7-32
srawi rTMP3, rTMP3, 3
add rRTN, rTMP1, rTMP3
blr
#endif
END (strlen)
libc_hidden_builtin_def (strlen)
|