1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006,
2 2008 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 in glibc by Roland McGrath (roland@ai.mit.edu).
9 Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
11 This program is free software: you can redistribute it and/or modify it
12 under the terms of the GNU General Public License as published by the
13 Free Software Foundation; either version 3 of the License, or any
16 This program is distributed in the hope that it will be useful,
17 but WITHOUT ANY WARRANTY; without even the implied warranty of
18 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
19 GNU General Public License for more details.
21 You should have received a copy of the GNU General Public License
22 along with this program. If not, see <http://www.gnu.org/licenses/>. */
32 /* Return the first address of either C1 or C2 (treated as unsigned
33 char) that occurs within N bytes of the memory region S. If
34 neither byte appears, return NULL. */
36 memchr2 (void const *s, int c1_in, int c2_in, size_t n)
38 const unsigned char *char_ptr;
39 const uintmax_t *longword_ptr;
49 c1 = (unsigned char) c1_in;
50 c2 = (unsigned char) c2_in;
53 return memchr (s, c1, n);
55 /* Handle the first few characters by reading one character at a time.
56 Do this until CHAR_PTR is aligned on a longword boundary. */
57 for (char_ptr = (const unsigned char *) s;
58 n > 0 && (size_t) char_ptr % sizeof longword1 != 0;
60 if (*char_ptr == c1 || *char_ptr == c2)
61 return (void *) char_ptr;
63 /* All these elucidatory comments refer to 4-byte longwords,
64 but the theory applies equally well to any size longwords. */
66 longword_ptr = (const uintmax_t *) char_ptr;
68 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
69 the "holes." Note that there is a hole just to the left of
70 each byte, with an extra at the end:
72 bits: 01111110 11111110 11111110 11111111
73 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
75 The 1-bits make sure that carries propagate to the next 0-bit.
76 The 0-bits provide holes for carries to fall into. */
78 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
79 Set CHARMASK to be a longword, each of whose bytes is C. */
81 magic_bits = 0xfefefefe;
82 charmask1 = c1 | (c1 << 8);
83 charmask2 = c2 | (c2 << 8);
84 charmask1 |= charmask1 << 16;
85 charmask2 |= charmask2 << 16;
86 if (0xffffffffU < UINTMAX_MAX)
88 magic_bits |= magic_bits << 32;
89 charmask1 |= charmask1 << 32;
90 charmask2 |= charmask2 << 32;
91 if (8 < sizeof longword1)
92 for (i = 64; i < sizeof longword1 * 8; i *= 2)
94 magic_bits |= magic_bits << i;
95 charmask1 |= charmask1 << i;
96 charmask2 |= charmask2 << i;
99 magic_bits = (UINTMAX_MAX >> 1) & (magic_bits | 1);
101 /* Instead of the traditional loop which tests each character,
102 we will test a longword at a time. The tricky part is testing
103 if *any of the four* bytes in the longword in question are zero. */
104 while (n >= sizeof longword1)
106 /* We tentatively exit the loop if adding MAGIC_BITS to
107 LONGWORD fails to change any of the hole bits of LONGWORD.
109 1) Is this safe? Will it catch all the zero bytes?
110 Suppose there is a byte with all zeros. Any carry bits
111 propagating from its left will fall into the hole at its
112 least significant bit and stop. Since there will be no
113 carry from its most significant bit, the LSB of the
114 byte to the left will be unchanged, and the zero will be
117 2) Is this worthwhile? Will it ignore everything except
118 zero bytes? Suppose every byte of LONGWORD has a bit set
119 somewhere. There will be a carry into bit 8. If bit 8
120 is set, this will carry into bit 16. If bit 8 is clear,
121 one of bits 9-15 must be set, so there will be a carry
122 into bit 16. Similarly, there will be a carry into bit
123 24. If one of bits 24-30 is set, there will be a carry
124 into bit 31, so all of the hole bits will be changed.
126 The one misfire occurs when bits 24-30 are clear and bit
127 31 is set; in this case, the hole at bit 31 is not
128 changed. If we had access to the processor carry flag,
129 we could close this loophole by putting the fourth hole
132 So it ignores everything except 128's, when they're aligned
135 3) But wait! Aren't we looking for C, not zero?
136 Good point. So what we do is XOR LONGWORD with a longword,
137 each of whose bytes is C. This turns each byte that is C
140 longword1 = *longword_ptr ^ charmask1;
141 longword2 = *longword_ptr++ ^ charmask2;
143 /* Add MAGIC_BITS to LONGWORD. */
144 if ((((longword1 + magic_bits)
146 /* Set those bits that were unchanged by the addition. */
149 /* Look at only the hole bits. If any of the hole bits
150 are unchanged, most likely one of the bytes was a
153 || (((longword2 + magic_bits) ^ ~longword2) & ~magic_bits) != 0)
155 /* Which of the bytes was C? If none of them were, it was
156 a misfire; continue the search. */
158 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
160 if (cp[0] == c1 || cp[0] == c2)
162 if (cp[1] == c1 || cp[1] == c2)
163 return (void *) &cp[1];
164 if (cp[2] == c1 || cp[2] == c2)
165 return (void *) &cp[2];
166 if (cp[3] == c1 || cp[3] == c2)
167 return (void *) &cp[3];
168 if (4 < sizeof longword1 && (cp[4] == c1 || cp[4] == c2))
169 return (void *) &cp[4];
170 if (5 < sizeof longword1 && (cp[5] == c1 || cp[5] == c2))
171 return (void *) &cp[5];
172 if (6 < sizeof longword1 && (cp[6] == c1 || cp[6] == c2))
173 return (void *) &cp[6];
174 if (7 < sizeof longword1 && (cp[7] == c1 || cp[7] == c2))
175 return (void *) &cp[7];
176 if (8 < sizeof longword1)
177 for (i = 8; i < sizeof longword1; i++)
178 if (cp[i] == c1 || cp[i] == c2)
179 return (void *) &cp[i];
182 n -= sizeof longword1;
185 char_ptr = (const unsigned char *) longword_ptr;
189 if (*char_ptr == c1 || *char_ptr == c2)
190 return (void *) char_ptr;