1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006, 2008
2 Free 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 3 of the License, or 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, see <http://www.gnu.org/licenses/>. */
37 # define reg_char char
42 #if HAVE_BP_SYM_H || defined _LIBC
45 # define BP_SYM(sym) sym
54 # define __memchr memchr
57 /* Search no more than N bytes of S for C. */
59 __memchr (void const *s, int c_in, size_t n)
61 const unsigned char *char_ptr;
62 const unsigned long int *longword_ptr;
63 unsigned long int longword, magic_bits, charmask;
67 c = (unsigned char) c_in;
69 /* Handle the first few characters by reading one character at a time.
70 Do this until CHAR_PTR is aligned on a longword boundary. */
71 for (char_ptr = (const unsigned char *) s;
72 n > 0 && (size_t) char_ptr % sizeof longword != 0;
75 return (void *) char_ptr;
77 /* All these elucidatory comments refer to 4-byte longwords,
78 but the theory applies equally well to any size longwords. */
80 longword_ptr = (const unsigned long int *) char_ptr;
82 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
83 the "holes." Note that there is a hole just to the left of
84 each byte, with an extra at the end:
86 bits: 01111110 11111110 11111110 11111111
87 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
89 The 1-bits make sure that carries propagate to the next 0-bit.
90 The 0-bits provide holes for carries to fall into. */
92 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
93 Set CHARMASK to be a longword, each of whose bytes is C. */
95 magic_bits = 0xfefefefe;
96 charmask = c | (c << 8);
97 charmask |= charmask << 16;
98 #if 0xffffffffU < ULONG_MAX
99 magic_bits |= magic_bits << 32;
100 charmask |= charmask << 32;
101 if (8 < sizeof longword)
102 for (i = 64; i < sizeof longword * 8; i *= 2)
104 magic_bits |= magic_bits << i;
105 charmask |= charmask << i;
108 magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
110 /* Instead of the traditional loop which tests each character,
111 we will test a longword at a time. The tricky part is testing
112 if *any of the four* bytes in the longword in question are zero. */
113 while (n >= sizeof longword)
115 /* We tentatively exit the loop if adding MAGIC_BITS to
116 LONGWORD fails to change any of the hole bits of LONGWORD.
118 1) Is this safe? Will it catch all the zero bytes?
119 Suppose there is a byte with all zeros. Any carry bits
120 propagating from its left will fall into the hole at its
121 least significant bit and stop. Since there will be no
122 carry from its most significant bit, the LSB of the
123 byte to the left will be unchanged, and the zero will be
126 2) Is this worthwhile? Will it ignore everything except
127 zero bytes? Suppose every byte of LONGWORD has a bit set
128 somewhere. There will be a carry into bit 8. If bit 8
129 is set, this will carry into bit 16. If bit 8 is clear,
130 one of bits 9-15 must be set, so there will be a carry
131 into bit 16. Similarly, there will be a carry into bit
132 24. If one of bits 24-30 is set, there will be a carry
133 into bit 31, so all of the hole bits will be changed.
135 The one misfire occurs when bits 24-30 are clear and bit
136 31 is set; in this case, the hole at bit 31 is not
137 changed. If we had access to the processor carry flag,
138 we could close this loophole by putting the fourth hole
141 So it ignores everything except 128's, when they're aligned
144 3) But wait! Aren't we looking for C, not zero?
145 Good point. So what we do is XOR LONGWORD with a longword,
146 each of whose bytes is C. This turns each byte that is C
149 longword = *longword_ptr++ ^ charmask;
151 /* Add MAGIC_BITS to LONGWORD. */
152 if ((((longword + magic_bits)
154 /* Set those bits that were unchanged by the addition. */
157 /* Look at only the hole bits. If any of the hole bits
158 are unchanged, most likely one of the bytes was a
162 /* Which of the bytes was C? If none of them were, it was
163 a misfire; continue the search. */
165 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
170 return (void *) &cp[1];
172 return (void *) &cp[2];
174 return (void *) &cp[3];
175 if (4 < sizeof longword && cp[4] == c)
176 return (void *) &cp[4];
177 if (5 < sizeof longword && cp[5] == c)
178 return (void *) &cp[5];
179 if (6 < sizeof longword && cp[6] == c)
180 return (void *) &cp[6];
181 if (7 < sizeof longword && cp[7] == c)
182 return (void *) &cp[7];
183 if (8 < sizeof longword)
184 for (i = 8; i < sizeof longword; i++)
186 return (void *) &cp[i];
189 n -= sizeof longword;
192 char_ptr = (const unsigned char *) longword_ptr;
197 return (void *) char_ptr;
205 weak_alias (__memchr, BP_SYM (memchr))