1 /* memchr (str, ch, n) -- Return pointer to first occurrence of CH in STR less
4 Copyright (C) 1994, 1995, 1996, 1997 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 Optimised a little by Alan Modra <Alan@SPRI.Levels.UniSA.Edu.Au>
9 This version is developed using the same algorithm as the fast C
10 version which carries the following introduction:
12 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
13 with help from Dan Sahlin (dan@sics.se) and
14 commentary by Jim Blandy (jimb@ai.mit.edu);
15 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
16 and implemented by Roland McGrath (roland@ai.mit.edu).
18 The GNU C Library is free software; you can redistribute it and/or
19 modify it under the terms of the GNU Library General Public License as
20 published by the Free Software Foundation; either version 2 of the
21 License, or (at your option) any later version.
23 The GNU C Library is distributed in the hope that it will be useful,
24 but WITHOUT ANY WARRANTY; without even the implied warranty of
25 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
26 Library General Public License for more details.
28 You should have received a copy of the GNU Library General Public
29 License along with the GNU C Library; see the file COPYING.LIB. If not,
30 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
31 Boston, MA 02111-1307, USA. */
34 #include "asm-syntax.h"
45 /* Save callee-safe registers used in this function. */
49 /* Load parameters into registers. */
50 movl 12(%esp), %eax /* str: pointer to memory block. */
51 movl 16(%esp), %edx /* c: byte we are looking for. */
52 movl 20(%esp), %esi /* len: length of memory block. */
54 /* If my must not test more than three characters test
55 them one by one. This is especially true for 0. */
59 /* At the moment %edx contains C. What we need for the
60 algorithm is C in all bytes of the dword. Avoid
61 operations on 16 bit words because these require an
62 prefix byte (and one more cycle). */
63 movb %dl, %dh /* Now it is 0|0|c|c */
65 shll $16, %edx /* Now c|c|0|0 */
66 movw %cx, %dx /* And finally c|c|c|c */
68 /* Better performance can be achieved if the word (32
69 bit) memory access is aligned on a four-byte-boundary.
70 So process first bytes one by one until boundary is
71 reached. Don't use a loop for better performance. */
73 testb $3, %eax /* correctly aligned ? */
74 je L(2) /* yes => begin loop */
75 cmpb %dl, (%eax) /* compare byte */
76 je L(9) /* target found => return */
77 incl %eax /* increment source pointer */
78 decl %esi /* decrement length counter */
79 je L(4) /* len==0 => return NULL */
81 testb $3, %eax /* correctly aligned ? */
82 je L(2) /* yes => begin loop */
83 cmpb %dl, (%eax) /* compare byte */
84 je L(9) /* target found => return */
85 incl %eax /* increment source pointer */
86 decl %esi /* decrement length counter */
87 je L(4) /* len==0 => return NULL */
89 testb $3, %eax /* correctly aligned ? */
90 je L(2) /* yes => begin loop */
91 cmpb %dl, (%eax) /* compare byte */
92 je L(9) /* target found => return */
93 incl %eax /* increment source pointer */
94 decl %esi /* decrement length counter */
95 /* no test for len==0 here, because this is done in the
99 /* We exit the loop if adding MAGIC_BITS to LONGWORD fails to
100 change any of the hole bits of LONGWORD.
102 1) Is this safe? Will it catch all the zero bytes?
103 Suppose there is a byte with all zeros. Any carry bits
104 propagating from its left will fall into the hole at its
105 least significant bit and stop. Since there will be no
106 carry from its most significant bit, the LSB of the
107 byte to the left will be unchanged, and the zero will be
110 2) Is this worthwhile? Will it ignore everything except
111 zero bytes? Suppose every byte of LONGWORD has a bit set
112 somewhere. There will be a carry into bit 8. If bit 8
113 is set, this will carry into bit 16. If bit 8 is clear,
114 one of bits 9-15 must be set, so there will be a carry
115 into bit 16. Similarly, there will be a carry into bit
116 24. If one of bits 24-31 is set, there will be a carry
117 into bit 32 (=carry flag), so all of the hole bits will
120 3) But wait! Aren't we looking for C, not zero?
121 Good point. So what we do is XOR LONGWORD with a longword,
122 each of whose bytes is C. This turns each byte that is C
126 /* Each round the main loop processes 16 bytes. */
130 L(1): movl (%eax), %ecx /* get word (= 4 bytes) in question */
131 movl $0xfefefeff, %edi /* magic value */
132 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
134 addl %ecx, %edi /* add the magic value to the word. We get
135 carry bits reported for each byte which
138 /* According to the algorithm we had to reverse the effect of the
139 XOR first and then test the overflow bits. But because the
140 following XOR would destroy the carry flag and it would (in a
141 representation with more than 32 bits) not alter then last
142 overflow, we can now test this condition. If no carry is signaled
143 no overflow must have occurred in the last byte => it was 0. */
146 /* We are only interested in carry bits that change due to the
147 previous add, so remove original bits */
148 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
150 /* Now test for the other three overflow bits. */
151 orl $0xfefefeff, %edi /* set all non-carry bits */
152 incl %edi /* add 1: if one carry bit was *not* set
153 the addition will not result in 0. */
155 /* If at least one byte of the word is C we don't get 0 in %edi. */
156 jnz L(8) /* found it => return pointer */
158 /* This process is unfolded four times for better performance.
159 we don't increment the source pointer each time. Instead we
160 use offsets and increment by 16 in each run of the loop. But
161 before probing for the matching byte we need some extra code
162 (following LL(13) below). Even the len can be compared with
163 constants instead of decrementing each time. */
165 movl 4(%eax), %ecx /* get word (= 4 bytes) in question */
166 movl $0xfefefeff, %edi /* magic value */
167 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
169 addl %ecx, %edi /* add the magic value to the word. We get
170 carry bits reported for each byte which
172 jnc L(7) /* 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(7) /* found it => return pointer */
179 movl 8(%eax), %ecx /* get word (= 4 bytes) in question */
180 movl $0xfefefeff, %edi /* magic value */
181 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
183 addl %ecx, %edi /* add the magic value to the word. We get
184 carry bits reported for each byte which
186 jnc L(6) /* highest byte is C => return pointer */
187 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
188 orl $0xfefefeff, %edi /* set all non-carry bits */
189 incl %edi /* add 1: if one carry bit was *not* set
190 the addition will not result in 0. */
191 jnz L(6) /* found it => return pointer */
193 movl 12(%eax), %ecx /* get word (= 4 bytes) in question */
194 movl $0xfefefeff, %edi /* magic value */
195 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
197 addl %ecx, %edi /* add the magic value to the word. We get
198 carry bits reported for each byte which
200 jnc L(5) /* highest byte is C => return pointer */
201 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
202 orl $0xfefefeff, %edi /* set all non-carry bits */
203 incl %edi /* add 1: if one carry bit was *not* set
204 the addition will not result in 0. */
205 jnz L(5) /* found it => return pointer */
207 /* Adjust both counters for a full round, i.e. 16 bytes. */
210 jae L(1) /* Still more than 16 bytes remaining */
212 /* Process remaining bytes separately. */
213 cmpl $4-16, %esi /* rest < 4 bytes? */
214 jb L(3) /* yes, than test byte by byte */
216 movl (%eax), %ecx /* get word (= 4 bytes) in question */
217 movl $0xfefefeff, %edi /* magic value */
218 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
220 addl %ecx, %edi /* add the magic value to the word. We get
221 carry bits reported for each byte which
223 jnc L(8) /* highest byte is C => return pointer */
224 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
225 orl $0xfefefeff, %edi /* set all non-carry bits */
226 incl %edi /* add 1: if one carry bit was *not* set
227 the addition will not result in 0. */
228 jne L(8) /* found it => return pointer */
229 addl $4, %eax /* adjust source pointer */
231 cmpl $8-16, %esi /* rest < 8 bytes? */
232 jb L(3) /* yes, than test byte by byte */
234 movl (%eax), %ecx /* get word (= 4 bytes) in question */
235 movl $0xfefefeff, %edi /* magic value */
236 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
238 addl %ecx, %edi /* add the magic value to the word. We get
239 carry bits reported for each byte which
241 jnc L(8) /* highest byte is C => return pointer */
242 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
243 orl $0xfefefeff, %edi /* set all non-carry bits */
244 incl %edi /* add 1: if one carry bit was *not* set
245 the addition will not result in 0. */
246 jne L(8) /* found it => return pointer */
247 addl $4, %eax /* adjust source pointer */
249 cmpl $12-16, %esi /* rest < 12 bytes? */
250 jb L(3) /* yes, than test byte by byte */
252 movl (%eax), %ecx /* get word (= 4 bytes) in question */
253 movl $0xfefefeff, %edi /* magic value */
254 xorl %edx, %ecx /* XOR with word c|c|c|c => bytes of str == c
256 addl %ecx, %edi /* add the magic value to the word. We get
257 carry bits reported for each byte which
259 jnc L(8) /* highest byte is C => return pointer */
260 xorl %ecx, %edi /* ((word^charmask)+magic)^(word^charmask) */
261 orl $0xfefefeff, %edi /* set all non-carry bits */
262 incl %edi /* add 1: if one carry bit was *not* set
263 the addition will not result in 0. */
264 jne L(8) /* found it => return pointer */
265 addl $4, %eax /* adjust source pointer */
267 /* Check the remaining bytes one by one. */
268 L(3): andl $3, %esi /* mask out uninteresting bytes */
269 jz L(4) /* no remaining bytes => return NULL */
271 cmpb %dl, (%eax) /* compare byte with C */
272 je L(9) /* equal, than return pointer */
273 incl %eax /* increment source pointer */
274 decl %esi /* decrement length */
275 jz L(4) /* no remaining bytes => return NULL */
277 cmpb %dl, (%eax) /* compare byte with C */
278 je L(9) /* equal, than return pointer */
279 incl %eax /* increment source pointer */
280 decl %esi /* decrement length */
281 jz L(4) /* no remaining bytes => return NULL */
283 cmpb %dl, (%eax) /* compare byte with C */
284 je L(9) /* equal, than return pointer */
286 L(4): /* no byte found => return NULL */
290 /* add missing source pointer increments */
295 /* Test for the matching byte in the word. %ecx contains a NUL
296 char in the byte which originally was the byte we are looking
298 L(8): testb %cl, %cl /* test first byte in dword */
299 jz L(9) /* if zero => return pointer */
300 incl %eax /* increment source pointer */
302 testb %ch, %ch /* test second byte in dword */
303 jz L(9) /* if zero => return pointer */
304 incl %eax /* increment source pointer */
306 testl $0xff0000, %ecx /* test third byte in dword */
307 jz L(9) /* if zero => return pointer */
308 incl %eax /* increment source pointer */
310 /* No further test needed we we know it is one of the four bytes. */
312 L(9): popl %edi /* pop saved registers */