1 /* strchr (str, ch) -- Return pointer to first occurrence of CH in STR.
3 Copyright (C) 1994-1997,1999,2000,2002,2003,2005
4 Free Software Foundation, Inc.
5 This file is part of the GNU C Library.
6 Contributed by Ulrich Drepper <drepper@gnu.ai.mit.edu>
7 Some optimisations by Alan Modra <Alan@SPRI.Levels.UniSA.Edu.Au>
9 The GNU C Library is free software; you can redistribute it and/or
10 modify it under the terms of the GNU Lesser General Public
11 License as published by the Free Software Foundation; either
12 version 2.1 of the License, or (at your option) any later version.
14 The GNU C Library is distributed in the hope that it will be useful,
15 but WITHOUT ANY WARRANTY; without even the implied warranty of
16 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
17 Lesser General Public License for more details.
19 You should have received a copy of the GNU Lesser General Public
20 License along with the GNU C Library; if not, write to the Free
21 Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
25 #include "asm-syntax.h"
29 #define PARMS LINKAGE+4 /* space for 1 saved reg */
31 #define STR RTN+RTN_SIZE
32 #define CHR STR+PTR_SIZE
35 ENTRY (BP_SYM (strchr))
38 pushl %edi /* Save callee-safe registers used here. */
39 cfi_adjust_cfa_offset (4)
40 cfi_rel_offset (edi, 0)
43 CHECK_BOUNDS_LOW (%eax, STR(%esp))
45 /* At the moment %edx contains C. What we need for the
46 algorithm is C in all bytes of the dword. Avoid
47 operations on 16 bit words because these require an
48 prefix byte (and one more cycle). */
49 movb %dl, %dh /* now it is 0|0|c|c */
51 shll $16, %edx /* now it is c|c|0|0 */
52 movw %cx, %dx /* and finally c|c|c|c */
54 /* Before we start with the main loop we process single bytes
55 until the source pointer is aligned. This has two reasons:
56 1. aligned 32-bit memory access is faster
58 2. we process in the main loop 32 bit in one step although
59 we don't know the end of the string. But accessing at
60 4-byte alignment guarantees that we never access illegal
61 memory if this would not also be done by the trivial
62 implementation (this is because all processor inherent
63 boundaries are multiples of 4. */
65 testb $3, %al /* correctly aligned ? */
66 jz L(11) /* yes => begin loop */
67 movb (%eax), %cl /* load byte in question (we need it twice) */
68 cmpb %cl, %dl /* compare byte */
69 je L(6) /* target found => return */
70 testb %cl, %cl /* is NUL? */
71 jz L(2) /* yes => return NULL */
72 incl %eax /* increment pointer */
74 testb $3, %al /* correctly aligned ? */
75 jz L(11) /* yes => begin loop */
76 movb (%eax), %cl /* load byte in question (we need it twice) */
77 cmpb %cl, %dl /* compare byte */
78 je L(6) /* target found => return */
79 testb %cl, %cl /* is NUL? */
80 jz L(2) /* yes => return NULL */
81 incl %eax /* increment pointer */
83 testb $3, %al /* correctly aligned ? */
84 jz L(11) /* yes => begin loop */
85 movb (%eax), %cl /* load byte in question (we need it twice) */
86 cmpb %cl, %dl /* compare byte */
87 je L(6) /* target found => return */
88 testb %cl, %cl /* is NUL? */
89 jz L(2) /* yes => return NULL */
90 incl %eax /* increment pointer */
92 /* No we have reached alignment. */
93 jmp L(11) /* begin loop */
95 /* We exit the loop if adding MAGIC_BITS to LONGWORD fails to
96 change any of the hole bits of LONGWORD.
98 1) Is this safe? Will it catch all the zero bytes?
99 Suppose there is a byte with all zeros. Any carry bits
100 propagating from its left will fall into the hole at its
101 least significant bit and stop. Since there will be no
102 carry from its most significant bit, the LSB of the
103 byte to the left will be unchanged, and the zero will be
106 2) Is this worthwhile? Will it ignore everything except
107 zero bytes? Suppose every byte of LONGWORD has a bit set
108 somewhere. There will be a carry into bit 8. If bit 8
109 is set, this will carry into bit 16. If bit 8 is clear,
110 one of bits 9-15 must be set, so there will be a carry
111 into bit 16. Similarly, there will be a carry into bit
112 24. If one of bits 24-31 is set, there will be a carry
113 into bit 32 (=carry flag), so all of the hole bits will
116 3) But wait! Aren't we looking for C, not zero?
117 Good point. So what we do is XOR LONGWORD with a longword,
118 each of whose bytes is C. This turns each byte that is C
121 /* Each round the main loop processes 16 bytes. */
125 L(1): addl $16, %eax /* adjust pointer for whole round */
127 L(11): movl (%eax), %ecx /* get word (= 4 bytes) in question */
128 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
130 movl $0xfefefeff, %edi /* magic value */
131 addl %ecx, %edi /* add the magic value to the word. We get
132 carry bits reported for each byte which
135 /* According to the algorithm we had to reverse the effect of the
136 XOR first and then test the overflow bits. But because the
137 following XOR would destroy the carry flag and it would (in a
138 representation with more than 32 bits) not alter then last
139 overflow, we can now test this condition. If no carry is signaled
140 no overflow must have occurred in the last byte => it was 0. */
143 /* We are only interested in carry bits that change due to the
144 previous add, so remove original bits */
145 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
147 /* Now test for the other three overflow bits. */
148 orl $0xfefefeff, %edi /* set all non-carry bits */
149 incl %edi /* add 1: if one carry bit was *not* set
150 the addition will not result in 0. */
152 /* If at least one byte of the word is C we don't get 0 in %edi. */
153 jnz L(7) /* found it => return pointer */
155 /* Now we made sure the dword does not contain the character we are
156 looking for. But because we deal with strings we have to check
157 for the end of string before testing the next dword. */
159 xorl %edx, %ecx /* restore original dword without reload */
160 movl $0xfefefeff, %edi /* magic value */
161 addl %ecx, %edi /* add the magic value to the word. We get
162 carry bits reported for each byte which
164 jnc L(2) /* highest byte is NUL => return NULL */
165 xorl %ecx, %edi /* (word+magic)^word */
166 orl $0xfefefeff, %edi /* set all non-carry bits */
167 incl %edi /* add 1: if one carry bit was *not* set
168 the addition will not result in 0. */
169 jnz L(2) /* found NUL => return NULL */
171 movl 4(%eax), %ecx /* get word (= 4 bytes) in question */
172 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
174 movl $0xfefefeff, %edi /* magic value */
175 addl %ecx, %edi /* add the magic value to the word. We get
176 carry bits reported for each byte which
178 jnc L(71) /* highest byte is C => return pointer */
179 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
180 orl $0xfefefeff, %edi /* set all non-carry bits */
181 incl %edi /* add 1: if one carry bit was *not* set
182 the addition will not result in 0. */
183 jnz L(71) /* found it => return pointer */
184 xorl %edx, %ecx /* restore original dword without reload */
185 movl $0xfefefeff, %edi /* magic value */
186 addl %ecx, %edi /* add the magic value to the word. We get
187 carry bits reported for each byte which
189 jnc L(2) /* highest byte is NUL => return NULL */
190 xorl %ecx, %edi /* (word+magic)^word */
191 orl $0xfefefeff, %edi /* set all non-carry bits */
192 incl %edi /* add 1: if one carry bit was *not* set
193 the addition will not result in 0. */
194 jnz L(2) /* found NUL => return NULL */
196 movl 8(%eax), %ecx /* get word (= 4 bytes) in question */
197 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
199 movl $0xfefefeff, %edi /* magic value */
200 addl %ecx, %edi /* add the magic value to the word. We get
201 carry bits reported for each byte which
203 jnc L(72) /* highest byte is C => return pointer */
204 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
205 orl $0xfefefeff, %edi /* set all non-carry bits */
206 incl %edi /* add 1: if one carry bit was *not* set
207 the addition will not result in 0. */
208 jnz L(72) /* found it => return pointer */
209 xorl %edx, %ecx /* restore original dword without reload */
210 movl $0xfefefeff, %edi /* magic value */
211 addl %ecx, %edi /* add the magic value to the word. We get
212 carry bits reported for each byte which
214 jnc L(2) /* highest byte is NUL => return NULL */
215 xorl %ecx, %edi /* (word+magic)^word */
216 orl $0xfefefeff, %edi /* set all non-carry bits */
217 incl %edi /* add 1: if one carry bit was *not* set
218 the addition will not result in 0. */
219 jnz L(2) /* found NUL => return NULL */
221 movl 12(%eax), %ecx /* get word (= 4 bytes) in question */
222 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
224 movl $0xfefefeff, %edi /* magic value */
225 addl %ecx, %edi /* add the magic value to the word. We get
226 carry bits reported for each byte which
228 jnc L(73) /* highest byte is C => return pointer */
229 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
230 orl $0xfefefeff, %edi /* set all non-carry bits */
231 incl %edi /* add 1: if one carry bit was *not* set
232 the addition will not result in 0. */
233 jnz L(73) /* found it => return pointer */
234 xorl %edx, %ecx /* restore original dword without reload */
235 movl $0xfefefeff, %edi /* magic value */
236 addl %ecx, %edi /* add the magic value to the word. We get
237 carry bits reported for each byte which
239 jnc L(2) /* highest byte is NUL => return NULL */
240 xorl %ecx, %edi /* (word+magic)^word */
241 orl $0xfefefeff, %edi /* set all non-carry bits */
242 incl %edi /* add 1: if one carry bit was *not* set
243 the addition will not result in 0. */
244 jz L(1) /* no NUL found => restart loop */
246 L(2): /* Return NULL. */
248 RETURN_NULL_BOUNDED_POINTER
249 popl %edi /* restore saved register content */
250 cfi_adjust_cfa_offset (-4)
256 cfi_adjust_cfa_offset (4)
257 cfi_rel_offset (edi, 0)
258 L(73): addl $4, %eax /* adjust pointer */
262 /* We now scan for the byte in which the character was matched.
263 But we have to take care of the case that a NUL char is
264 found before this in the dword. Note that we XORed %ecx
265 with the byte we're looking for, therefore the tests below look
268 L(7): testb %cl, %cl /* is first byte C? */
269 jz L(6) /* yes => return pointer */
270 cmpb %dl, %cl /* is first byte NUL? */
271 je L(2) /* yes => return NULL */
272 incl %eax /* it's not in the first byte */
274 testb %ch, %ch /* is second byte C? */
275 jz L(6) /* yes => return pointer */
276 cmpb %dl, %ch /* is second byte NUL? */
277 je L(2) /* yes => return NULL? */
278 incl %eax /* it's not in the second byte */
280 shrl $16, %ecx /* make upper byte accessible */
281 testb %cl, %cl /* is third byte C? */
282 jz L(6) /* yes => return pointer */
283 cmpb %dl, %cl /* is third byte NUL? */
284 je L(2) /* yes => return NULL */
286 /* It must be in the fourth byte and it cannot be NUL. */
290 CHECK_BOUNDS_HIGH (%eax, STR(%esp), jb)
291 RETURN_BOUNDED_POINTER (STR(%esp))
292 popl %edi /* restore saved register content */
293 cfi_adjust_cfa_offset (-4)
298 END (BP_SYM (strchr))
300 weak_alias (BP_SYM (strchr), BP_SYM (index))
301 libc_hidden_builtin_def (strchr)