1 /* strchr (str, ch) -- Return pointer to first occurrence of CH in STR.
3 Copyright (C) 1994, 1995, 1996, 1997 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 Some optimisations by Alan Modra <Alan@SPRI.Levels.UniSA.Edu.Au>
8 The GNU C Library is free software; you can redistribute it and/or
9 modify it under the terms of the GNU Library General Public License as
10 published by the Free Software Foundation; either version 2 of the
11 License, or (at your option) any later version.
13 The GNU C Library is distributed in the hope that it will be useful,
14 but WITHOUT ANY WARRANTY; without even the implied warranty of
15 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
16 Library General Public License for more details.
18 You should have received a copy of the GNU Library General Public
19 License along with the GNU C Library; see the file COPYING.LIB. If not,
20 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
21 Boston, MA 02111-1307, USA. */
24 #include "asm-syntax.h"
34 pushl %edi /* Save callee-safe registers used here. */
36 movl 8(%esp), %eax /* get string pointer */
37 movl 12(%esp), %edx /* get character we are looking for */
39 /* At the moment %edx contains C. What we need for the
40 algorithm is C in all bytes of the dword. Avoid
41 operations on 16 bit words because these require an
42 prefix byte (and one more cycle). */
43 movb %dl, %dh /* now it is 0|0|c|c */
45 shll $16, %edx /* now it is c|c|0|0 */
46 movw %cx, %dx /* and finally c|c|c|c */
48 /* Before we start with the main loop we process single bytes
49 until the source pointer is aligned. This has two reasons:
50 1. aligned 32-bit memory access is faster
52 2. we process in the main loop 32 bit in one step although
53 we don't know the end of the string. But accessing at
54 4-byte alignment guarantees that we never access illegal
55 memory if this would not also be done by the trivial
56 implementation (this is because all processor inherent
57 boundaries are multiples of 4. */
59 testb $3, %eax /* correctly aligned ? */
60 jz L(11) /* yes => begin loop */
61 movb (%eax), %cl /* load byte in question (we need it twice) */
62 cmpb %cl, %dl /* compare byte */
63 je L(6) /* target found => return */
64 testb %cl, %cl /* is NUL? */
65 jz L(2) /* yes => return NULL */
66 incl %eax /* increment pointer */
68 testb $3, %eax /* correctly aligned ? */
69 jz L(11) /* yes => begin loop */
70 movb (%eax), %cl /* load byte in question (we need it twice) */
71 cmpb %cl, %dl /* compare byte */
72 je L(6) /* target found => return */
73 testb %cl, %cl /* is NUL? */
74 jz L(2) /* yes => return NULL */
75 incl %eax /* increment pointer */
77 testb $3, %eax /* correctly aligned ? */
78 jz L(11) /* yes => begin loop */
79 movb (%eax), %cl /* load byte in question (we need it twice) */
80 cmpb %cl, %dl /* compare byte */
81 je L(6) /* target found => return */
82 testb %cl, %cl /* is NUL? */
83 jz L(2) /* yes => return NULL */
84 incl %eax /* increment pointer */
86 /* No we have reached alignment. */
87 jmp L(11) /* begin loop */
89 /* We exit the loop if adding MAGIC_BITS to LONGWORD fails to
90 change any of the hole bits of LONGWORD.
92 1) Is this safe? Will it catch all the zero bytes?
93 Suppose there is a byte with all zeros. Any carry bits
94 propagating from its left will fall into the hole at its
95 least significant bit and stop. Since there will be no
96 carry from its most significant bit, the LSB of the
97 byte to the left will be unchanged, and the zero will be
100 2) Is this worthwhile? Will it ignore everything except
101 zero bytes? Suppose every byte of LONGWORD has a bit set
102 somewhere. There will be a carry into bit 8. If bit 8
103 is set, this will carry into bit 16. If bit 8 is clear,
104 one of bits 9-15 must be set, so there will be a carry
105 into bit 16. Similarly, there will be a carry into bit
106 24. If one of bits 24-31 is set, there will be a carry
107 into bit 32 (=carry flag), so all of the hole bits will
110 3) But wait! Aren't we looking for C, not zero?
111 Good point. So what we do is XOR LONGWORD with a longword,
112 each of whose bytes is C. This turns each byte that is C
115 /* Each round the main loop processes 16 bytes. */
119 L(1): addl $16, %eax /* adjust pointer for whole round */
121 L(11): movl (%eax), %ecx /* get word (= 4 bytes) in question */
122 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
124 movl $0xfefefeff, %edi /* magic value */
125 addl %ecx, %edi /* add the magic value to the word. We get
126 carry bits reported for each byte which
129 /* According to the algorithm we had to reverse the effect of the
130 XOR first and then test the overflow bits. But because the
131 following XOR would destroy the carry flag and it would (in a
132 representation with more than 32 bits) not alter then last
133 overflow, we can now test this condition. If no carry is signaled
134 no overflow must have occurred in the last byte => it was 0. */
137 /* We are only interested in carry bits that change due to the
138 previous add, so remove original bits */
139 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
141 /* Now test for the other three overflow bits. */
142 orl $0xfefefeff, %edi /* set all non-carry bits */
143 incl %edi /* add 1: if one carry bit was *not* set
144 the addition will not result in 0. */
146 /* If at least one byte of the word is C we don't get 0 in %edi. */
147 jnz L(7) /* found it => return pointer */
149 /* Now we made sure the dword does not contain the character we are
150 looking for. But because we deal with strings we have to check
151 for the end of string before testing the next dword. */
153 xorl %edx, %ecx /* restore original dword without reload */
154 movl $0xfefefeff, %edi /* magic value */
155 addl %ecx, %edi /* add the magic value to the word. We get
156 carry bits reported for each byte which
158 jnc L(2) /* highest byte is NUL => return NULL */
159 xorl %ecx, %edi /* (word+magic)^word */
160 orl $0xfefefeff, %edi /* set all non-carry bits */
161 incl %edi /* add 1: if one carry bit was *not* set
162 the addition will not result in 0. */
163 jnz L(2) /* found NUL => return NULL */
165 movl 4(%eax), %ecx /* get word (= 4 bytes) in question */
166 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
168 movl $0xfefefeff, %edi /* magic value */
169 addl %ecx, %edi /* add the magic value to the word. We get
170 carry bits reported for each byte which
172 jnc L(71) /* highest byte is C => return pointer */
173 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
174 orl $0xfefefeff, %edi /* set all non-carry bits */
175 incl %edi /* add 1: if one carry bit was *not* set
176 the addition will not result in 0. */
177 jnz L(71) /* found it => return pointer */
178 xorl %edx, %ecx /* restore original dword without reload */
179 movl $0xfefefeff, %edi /* magic value */
180 addl %ecx, %edi /* add the magic value to the word. We get
181 carry bits reported for each byte which
183 jnc L(2) /* highest byte is NUL => return NULL */
184 xorl %ecx, %edi /* (word+magic)^word */
185 orl $0xfefefeff, %edi /* set all non-carry bits */
186 incl %edi /* add 1: if one carry bit was *not* set
187 the addition will not result in 0. */
188 jnz L(2) /* found NUL => return NULL */
190 movl 8(%eax), %ecx /* get word (= 4 bytes) in question */
191 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
193 movl $0xfefefeff, %edi /* magic value */
194 addl %ecx, %edi /* add the magic value to the word. We get
195 carry bits reported for each byte which
197 jnc L(72) /* highest byte is C => return pointer */
198 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
199 orl $0xfefefeff, %edi /* set all non-carry bits */
200 incl %edi /* add 1: if one carry bit was *not* set
201 the addition will not result in 0. */
202 jnz L(72) /* found it => return pointer */
203 xorl %edx, %ecx /* restore original dword without reload */
204 movl $0xfefefeff, %edi /* magic value */
205 addl %ecx, %edi /* add the magic value to the word. We get
206 carry bits reported for each byte which
208 jnc L(2) /* highest byte is NUL => return NULL */
209 xorl %ecx, %edi /* (word+magic)^word */
210 orl $0xfefefeff, %edi /* set all non-carry bits */
211 incl %edi /* add 1: if one carry bit was *not* set
212 the addition will not result in 0. */
213 jnz L(2) /* found NUL => return NULL */
215 movl 12(%eax), %ecx /* get word (= 4 bytes) in question */
216 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
218 movl $0xfefefeff, %edi /* magic value */
219 addl %ecx, %edi /* add the magic value to the word. We get
220 carry bits reported for each byte which
222 jnc L(73) /* highest byte is C => return pointer */
223 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
224 orl $0xfefefeff, %edi /* set all non-carry bits */
225 incl %edi /* add 1: if one carry bit was *not* set
226 the addition will not result in 0. */
227 jnz L(73) /* found it => return pointer */
228 xorl %edx, %ecx /* restore original dword without reload */
229 movl $0xfefefeff, %edi /* magic value */
230 addl %ecx, %edi /* add the magic value to the word. We get
231 carry bits reported for each byte which
233 jnc L(2) /* highest byte is NUL => return NULL */
234 xorl %ecx, %edi /* (word+magic)^word */
235 orl $0xfefefeff, %edi /* set all non-carry bits */
236 incl %edi /* add 1: if one carry bit was *not* set
237 the addition will not result in 0. */
238 jz L(1) /* no NUL found => restart loop */
240 L(2): /* Return NULL. */
241 xorl %eax, %eax /* load NULL in return value register */
242 popl %edi /* restore saved register content */
245 L(73): addl $4, %eax /* adjust pointer */
249 /* We now scan for the byte in which the character was matched.
250 But we have to take care of the case that a NUL char is
251 found before this in the dword. */
253 L(7): testb %cl, %cl /* is first byte C? */
254 jz L(6) /* yes => return pointer */
255 cmpb %dl, %cl /* is first byte NUL? */
256 je L(2) /* yes => return NULL */
257 incl %eax /* it's not in the first byte */
259 testb %ch, %ch /* is second byte C? */
260 jz L(6) /* yes => return pointer */
261 cmpb %dl, %ch /* is second byte NUL? */
262 je L(2) /* yes => return NULL? */
263 incl %eax /* it's not in the second byte */
265 shrl $16, %ecx /* make upper byte accessible */
266 testb %cl, %cl /* is third byte C? */
267 jz L(6) /* yes => return pointer */
268 cmpb %dl, %cl /* is third byte NUL? */
269 je L(2) /* yes => return NULL */
271 /* It must be in the fourth byte and it cannot be NUL. */
274 L(6): popl %edi /* restore saved register content */
279 weak_alias (strchr, index)