1 /* rawmemchr (str, ch) -- Return pointer to first occurrence of CH in STR.
3 Copyright (C) 1994-2014 Free Software Foundation, Inc.
4 This file is part of the GNU C Library.
5 Contributed by Ulrich Drepper <drepper@gnu.ai.mit.edu>
6 Optimised a little by Alan Modra <Alan@SPRI.Levels.UniSA.Edu.Au>
7 This version is developed using the same algorithm as the fast C
8 version which carries the following introduction:
9 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
10 with help from Dan Sahlin (dan@sics.se) and
11 commentary by Jim Blandy (jimb@ai.mit.edu);
12 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
13 and implemented by Roland McGrath (roland@ai.mit.edu).
15 The GNU C Library is free software; you can redistribute it and/or
16 modify it under the terms of the GNU Lesser General Public
17 License as published by the Free Software Foundation; either
18 version 2.1 of the License, or (at your option) any later version.
20 The GNU C Library is distributed in the hope that it will be useful,
21 but WITHOUT ANY WARRANTY; without even the implied warranty of
22 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
23 Lesser General Public License for more details.
25 You should have received a copy of the GNU Lesser General Public
26 License along with the GNU C Library; if not, see
27 <http://www.gnu.org/licenses/>. */
30 #include "asm-syntax.h"
32 #define PARMS 4+4 /* space for 1 saved reg */
40 /* Save callee-safe register used in this function. */
42 cfi_adjust_cfa_offset (4)
43 cfi_rel_offset (edi, 0)
45 /* Load parameters into registers. */
49 /* At the moment %edx contains C. What we need for the
50 algorithm is C in all bytes of the dword. Avoid
51 operations on 16 bit words because these require an
52 prefix byte (and one more cycle). */
53 movb %dl, %dh /* Now it is 0|0|c|c */
55 shll $16, %edx /* Now c|c|0|0 */
56 movw %cx, %dx /* And finally c|c|c|c */
58 /* Better performance can be achieved if the word (32
59 bit) memory access is aligned on a four-byte-boundary.
60 So process first bytes one by one until boundary is
61 reached. Don't use a loop for better performance. */
63 testb $3, %al /* correctly aligned ? */
64 je L(1) /* yes => begin loop */
65 cmpb %dl, (%eax) /* compare byte */
66 je L(9) /* target found => return */
67 incl %eax /* increment source pointer */
69 testb $3, %al /* correctly aligned ? */
70 je L(1) /* yes => begin loop */
71 cmpb %dl, (%eax) /* compare byte */
72 je L(9) /* target found => return */
73 incl %eax /* increment source pointer */
75 testb $3, %al /* correctly aligned ? */
76 je L(1) /* yes => begin loop */
77 cmpb %dl, (%eax) /* compare byte */
78 je L(9) /* target found => return */
79 incl %eax /* increment source pointer */
81 /* We exit the loop if adding MAGIC_BITS to LONGWORD fails to
82 change any of the hole bits of LONGWORD.
84 1) Is this safe? Will it catch all the zero bytes?
85 Suppose there is a byte with all zeros. Any carry bits
86 propagating from its left will fall into the hole at its
87 least significant bit and stop. Since there will be no
88 carry from its most significant bit, the LSB of the
89 byte to the left will be unchanged, and the zero will be
92 2) Is this worthwhile? Will it ignore everything except
93 zero bytes? Suppose every byte of LONGWORD has a bit set
94 somewhere. There will be a carry into bit 8. If bit 8
95 is set, this will carry into bit 16. If bit 8 is clear,
96 one of bits 9-15 must be set, so there will be a carry
97 into bit 16. Similarly, there will be a carry into bit
98 24. If one of bits 24-31 is set, there will be a carry
99 into bit 32 (=carry flag), so all of the hole bits will
102 3) But wait! Aren't we looking for C, not zero?
103 Good point. So what we do is XOR LONGWORD with a longword,
104 each of whose bytes is C. This turns each byte that is C
108 /* Each round the main loop processes 16 bytes. */
111 L(1): movl (%eax), %ecx /* get word (= 4 bytes) in question */
112 movl $0xfefefeff, %edi /* magic value */
113 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
115 addl %ecx, %edi /* add the magic value to the word. We get
116 carry bits reported for each byte which
119 /* According to the algorithm we had to reverse the effect of the
120 XOR first and then test the overflow bits. But because the
121 following XOR would destroy the carry flag and it would (in a
122 representation with more than 32 bits) not alter then last
123 overflow, we can now test this condition. If no carry is signaled
124 no overflow must have occurred in the last byte => it was 0. */
127 /* We are only interested in carry bits that change due to the
128 previous add, so remove original bits */
129 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
131 /* Now test for the other three overflow bits. */
132 orl $0xfefefeff, %edi /* set all non-carry bits */
133 incl %edi /* add 1: if one carry bit was *not* set
134 the addition will not result in 0. */
136 /* If at least one byte of the word is C we don't get 0 in %edi. */
137 jnz L(8) /* found it => return pointer */
139 /* This process is unfolded four times for better performance.
140 we don't increment the source pointer each time. Instead we
141 use offsets and increment by 16 in each run of the loop. But
142 before probing for the matching byte we need some extra code
143 (following LL(13) below). Even the len can be compared with
144 constants instead of decrementing each time. */
146 movl 4(%eax), %ecx /* get word (= 4 bytes) in question */
147 movl $0xfefefeff, %edi /* magic value */
148 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
150 addl %ecx, %edi /* add the magic value to the word. We get
151 carry bits reported for each byte which
153 jnc L(7) /* highest byte is C => return pointer */
154 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
155 orl $0xfefefeff, %edi /* set all non-carry bits */
156 incl %edi /* add 1: if one carry bit was *not* set
157 the addition will not result in 0. */
158 jnz L(7) /* found it => return pointer */
160 movl 8(%eax), %ecx /* get word (= 4 bytes) in question */
161 movl $0xfefefeff, %edi /* magic value */
162 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
164 addl %ecx, %edi /* add the magic value to the word. We get
165 carry bits reported for each byte which
167 jnc L(6) /* highest byte is C => return pointer */
168 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
169 orl $0xfefefeff, %edi /* set all non-carry bits */
170 incl %edi /* add 1: if one carry bit was *not* set
171 the addition will not result in 0. */
172 jnz L(6) /* found it => return pointer */
174 movl 12(%eax), %ecx /* get word (= 4 bytes) in question */
175 movl $0xfefefeff, %edi /* magic value */
176 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
178 addl %ecx, %edi /* add the magic value to the word. We get
179 carry bits reported for each byte which
181 jnc L(5) /* highest byte is C => return pointer */
182 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
183 orl $0xfefefeff, %edi /* set all non-carry bits */
184 incl %edi /* add 1: if one carry bit was *not* set
185 the addition will not result in 0. */
186 jnz L(5) /* found it => return pointer */
188 /* Adjust both counters for a full round, i.e. 16 bytes. */
191 /* add missing source pointer increments */
196 /* Test for the matching byte in the word. %ecx contains a NUL
197 char in the byte which originally was the byte we are looking
199 L(8): testb %cl, %cl /* test first byte in dword */
200 jz L(9) /* if zero => return pointer */
201 incl %eax /* increment source pointer */
203 testb %ch, %ch /* test second byte in dword */
204 jz L(9) /* if zero => return pointer */
205 incl %eax /* increment source pointer */
207 testl $0xff0000, %ecx /* test third byte in dword */
208 jz L(9) /* if zero => return pointer */
209 incl %eax /* increment source pointer */
211 /* No further test needed we we know it is one of the four bytes. */
214 popl %edi /* pop saved register */
215 cfi_adjust_cfa_offset (-4)
221 libc_hidden_def (__rawmemchr)
222 weak_alias (__rawmemchr, rawmemchr)