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 |= charmask2 << 16;
85 charmask1 |= charmask2 << 16;
86 #if 0xffffffffU < UINTMAX_MAX
87 magic_bits |= magic_bits << 32;
88 charmask1 |= charmask1 << 32;
89 charmask2 |= charmask2 << 32;
90 if (8 < sizeof longword1)
91 for (i = 64; i < sizeof longword1 * 8; i *= 2)
93 magic_bits |= magic_bits << i;
94 charmask1 |= charmask1 << i;
95 charmask2 |= charmask2 << i;
98 magic_bits = (UINTMAX_MAX >> 1) & (magic_bits | 1);
100 /* Instead of the traditional loop which tests each character,
101 we will test a longword at a time. The tricky part is testing
102 if *any of the four* bytes in the longword in question are zero. */
103 while (n >= sizeof longword1)
105 /* We tentatively exit the loop if adding MAGIC_BITS to
106 LONGWORD fails to change any of the hole bits of LONGWORD.
108 1) Is this safe? Will it catch all the zero bytes?
109 Suppose there is a byte with all zeros. Any carry bits
110 propagating from its left will fall into the hole at its
111 least significant bit and stop. Since there will be no
112 carry from its most significant bit, the LSB of the
113 byte to the left will be unchanged, and the zero will be
116 2) Is this worthwhile? Will it ignore everything except
117 zero bytes? Suppose every byte of LONGWORD has a bit set
118 somewhere. There will be a carry into bit 8. If bit 8
119 is set, this will carry into bit 16. If bit 8 is clear,
120 one of bits 9-15 must be set, so there will be a carry
121 into bit 16. Similarly, there will be a carry into bit
122 24. If one of bits 24-30 is set, there will be a carry
123 into bit 31, so all of the hole bits will be changed.
125 The one misfire occurs when bits 24-30 are clear and bit
126 31 is set; in this case, the hole at bit 31 is not
127 changed. If we had access to the processor carry flag,
128 we could close this loophole by putting the fourth hole
131 So it ignores everything except 128's, when they're aligned
134 3) But wait! Aren't we looking for C, not zero?
135 Good point. So what we do is XOR LONGWORD with a longword,
136 each of whose bytes is C. This turns each byte that is C
139 longword1 = *longword_ptr ^ charmask1;
140 longword2 = *longword_ptr++ ^ charmask2;
142 /* Add MAGIC_BITS to LONGWORD. */
143 if ((((longword1 + magic_bits)
145 /* Set those bits that were unchanged by the addition. */
148 /* Look at only the hole bits. If any of the hole bits
149 are unchanged, most likely one of the bytes was a
152 || (((longword2 + magic_bits) ^ ~longword2) & ~magic_bits) != 0)
154 /* Which of the bytes was C? If none of them were, it was
155 a misfire; continue the search. */
157 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
159 if (cp[0] == c1 || cp[0] == c2)
161 if (cp[1] == c1 || cp[1] == c2)
162 return (void *) &cp[1];
163 if (cp[2] == c1 || cp[2] == c2)
164 return (void *) &cp[2];
165 if (cp[3] == c1 || cp[3] == c2)
166 return (void *) &cp[3];
167 if (4 < sizeof longword1 && (cp[4] == c1 || cp[4] == c2))
168 return (void *) &cp[4];
169 if (5 < sizeof longword1 && (cp[5] == c1 || cp[5] == c2))
170 return (void *) &cp[5];
171 if (6 < sizeof longword1 && (cp[6] == c1 || cp[6] == c2))
172 return (void *) &cp[6];
173 if (7 < sizeof longword1 && (cp[7] == c1 || cp[7] == c2))
174 return (void *) &cp[7];
175 if (8 < sizeof longword1)
176 for (i = 8; i < sizeof longword1; i++)
177 if (cp[i] == c1 || cp[i] == c2)
178 return (void *) &cp[i];
181 n -= sizeof longword1;
184 char_ptr = (const unsigned char *) longword_ptr;
188 if (*char_ptr == c1 || *char_ptr == c2)
189 return (void *) char_ptr;