1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003 Free
2 Software Foundation, Inc.
4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
5 with help from Dan Sahlin (dan@sics.se) and
6 commentary by Jim Blandy (jimb@ai.mit.edu);
7 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
8 and implemented by Roland McGrath (roland@ai.mit.edu).
10 NOTE: The canonical source of this file is maintained with the GNU C Library.
11 Bugs can be reported to bug-glibc@prep.ai.mit.edu.
13 This program is free software; you can redistribute it and/or modify it
14 under the terms of the GNU General Public License as published by the
15 Free Software Foundation; either version 2, or (at your option) any
18 This program is distributed in the hope that it will be useful,
19 but WITHOUT ANY WARRANTY; without even the implied warranty of
20 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
21 GNU General Public License for more details.
23 You should have received a copy of the GNU General Public License
24 along with this program; if not, write to the Free Software
25 Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307,
37 # define reg_char char
43 #define LONG_MAX_32_BITS 2147483647
45 #include <sys/types.h>
46 #if HAVE_BP_SYM_H || defined _LIBC
49 # define BP_SYM(sym) sym
55 /* Search no more than N bytes of S for C. */
57 __memchr (void const *s, int c_in, size_t n)
59 const unsigned char *char_ptr;
60 const unsigned long int *longword_ptr;
61 unsigned long int longword, magic_bits, charmask;
64 c = (unsigned char) c_in;
66 /* Handle the first few characters by reading one character at a time.
67 Do this until CHAR_PTR is aligned on a longword boundary. */
68 for (char_ptr = (const unsigned char *) s;
69 n > 0 && ((unsigned long int) char_ptr
70 & (sizeof (longword) - 1)) != 0;
73 return (void *) char_ptr;
75 /* All these elucidatory comments refer to 4-byte longwords,
76 but the theory applies equally well to 8-byte longwords. */
78 longword_ptr = (unsigned long int *) char_ptr;
80 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
81 the "holes." Note that there is a hole just to the left of
82 each byte, with an extra at the end:
84 bits: 01111110 11111110 11111110 11111111
85 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
87 The 1-bits make sure that carries propagate to the next 0-bit.
88 The 0-bits provide holes for carries to fall into. */
90 if (sizeof (longword) != 4 && sizeof (longword) != 8)
93 #if LONG_MAX <= LONG_MAX_32_BITS
94 magic_bits = 0x7efefeff;
96 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
99 /* Set up a longword, each of whose bytes is C. */
100 charmask = c | (c << 8);
101 charmask |= charmask << 16;
102 #if LONG_MAX > LONG_MAX_32_BITS
103 charmask |= charmask << 32;
106 /* Instead of the traditional loop which tests each character,
107 we will test a longword at a time. The tricky part is testing
108 if *any of the four* bytes in the longword in question are zero. */
109 while (n >= sizeof (longword))
111 /* We tentatively exit the loop if adding MAGIC_BITS to
112 LONGWORD fails to change any of the hole bits of LONGWORD.
114 1) Is this safe? Will it catch all the zero bytes?
115 Suppose there is a byte with all zeros. Any carry bits
116 propagating from its left will fall into the hole at its
117 least significant bit and stop. Since there will be no
118 carry from its most significant bit, the LSB of the
119 byte to the left will be unchanged, and the zero will be
122 2) Is this worthwhile? Will it ignore everything except
123 zero bytes? Suppose every byte of LONGWORD has a bit set
124 somewhere. There will be a carry into bit 8. If bit 8
125 is set, this will carry into bit 16. If bit 8 is clear,
126 one of bits 9-15 must be set, so there will be a carry
127 into bit 16. Similarly, there will be a carry into bit
128 24. If one of bits 24-30 is set, there will be a carry
129 into bit 31, so all of the hole bits will be changed.
131 The one misfire occurs when bits 24-30 are clear and bit
132 31 is set; in this case, the hole at bit 31 is not
133 changed. If we had access to the processor carry flag,
134 we could close this loophole by putting the fourth hole
137 So it ignores everything except 128's, when they're aligned
140 3) But wait! Aren't we looking for C, not zero?
141 Good point. So what we do is XOR LONGWORD with a longword,
142 each of whose bytes is C. This turns each byte that is C
145 longword = *longword_ptr++ ^ charmask;
147 /* Add MAGIC_BITS to LONGWORD. */
148 if ((((longword + magic_bits)
150 /* Set those bits that were unchanged by the addition. */
153 /* Look at only the hole bits. If any of the hole bits
154 are unchanged, most likely one of the bytes was a
158 /* Which of the bytes was C? If none of them were, it was
159 a misfire; continue the search. */
161 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
166 return (void *) &cp[1];
168 return (void *) &cp[2];
170 return (void *) &cp[3];
171 #if LONG_MAX > 2147483647
173 return (void *) &cp[4];
175 return (void *) &cp[5];
177 return (void *) &cp[6];
179 return (void *) &cp[7];
183 n -= sizeof (longword);
186 char_ptr = (const unsigned char *) longword_ptr;
191 return (void *) char_ptr;
199 weak_alias (__memchr, BP_SYM (memchr))