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/>. */
34 /* Return the first address of either C1 or C2 (treated as unsigned
35 char) that occurs within N bytes of the memory region S. If
36 neither byte appears, return NULL. */
38 memchr2 (void const *s, int c1_in, int c2_in, size_t n)
40 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
41 long instead of a 64-bit uintmax_t tends to give better
42 performance. On 64-bit hardware, unsigned long is generally 64
43 bits already. Change this typedef to experiment with
45 typedef unsigned long longword;
47 const unsigned char *char_ptr;
48 const longword *longword_ptr;
49 longword repeated_one;
55 c1 = (unsigned char) c1_in;
56 c2 = (unsigned char) c2_in;
59 return memchr (s, c1, n);
61 /* Handle the first few bytes by reading one byte at a time.
62 Do this until CHAR_PTR is aligned on a longword boundary. */
63 for (char_ptr = (const unsigned char *) s;
64 n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
66 if (*char_ptr == c1 || *char_ptr == c2)
67 return (void *) char_ptr;
69 longword_ptr = (const longword *) char_ptr;
71 /* All these elucidatory comments refer to 4-byte longwords,
72 but the theory applies equally well to any size longwords. */
74 /* Compute auxiliary longword values:
75 repeated_one is a value which has a 1 in every byte.
76 repeated_c1 has c1 in every byte.
77 repeated_c2 has c2 in every byte. */
78 repeated_one = 0x01010101;
79 repeated_c1 = c1 | (c1 << 8);
80 repeated_c2 = c2 | (c2 << 8);
81 repeated_c1 |= repeated_c1 << 16;
82 repeated_c2 |= repeated_c2 << 16;
83 if (0xffffffffU < TYPE_MAXIMUM (longword))
85 repeated_one |= repeated_one << 31 << 1;
86 repeated_c1 |= repeated_c1 << 31 << 1;
87 repeated_c2 |= repeated_c2 << 31 << 1;
88 if (8 < sizeof (longword))
92 for (i = 64; i < sizeof (longword) * 8; i *= 2)
94 repeated_one |= repeated_one << i;
95 repeated_c1 |= repeated_c1 << i;
96 repeated_c2 |= repeated_c2 << i;
101 /* Instead of the traditional loop which tests each byte, we will test a
102 longword at a time. The tricky part is testing if *any of the four*
103 bytes in the longword in question are equal to c1 or c2. We first use
104 an xor with repeated_c1 and repeated_c2, respectively. This reduces
105 the task to testing whether *any of the four* bytes in longword1 or
108 Let's consider longword1. We compute tmp1 =
109 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
110 That is, we perform the following operations:
111 1. Subtract repeated_one.
113 3. & a mask consisting of 0x80 in every byte.
114 Consider what happens in each byte:
115 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
116 and step 3 transforms it into 0x80. A carry can also be propagated
117 to more significant bytes.
118 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
119 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
120 the byte ends in a single bit of value 0 and k bits of value 1.
121 After step 2, the result is just k bits of value 1: 2^k - 1. After
122 step 3, the result is 0. And no carry is produced.
123 So, if longword1 has only non-zero bytes, tmp1 is zero.
124 Whereas if longword1 has a zero byte, call j the position of the least
125 significant zero byte. Then the result has a zero at positions 0, ...,
126 j-1 and a 0x80 at position j. We cannot predict the result at the more
127 significant bytes (positions j+1..3), but it does not matter since we
128 already have a non-zero bit at position 8*j+7.
130 Similary, we compute tmp2 =
131 ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
133 The test whether any byte in longword1 or longword2 is zero is equivalent
134 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
135 this into a single test, whether (tmp1 | tmp2) is nonzero. */
137 while (n >= sizeof (longword))
139 longword longword1 = *longword_ptr ^ repeated_c1;
140 longword longword2 = *longword_ptr ^ repeated_c2;
142 if (((((longword1 - repeated_one) & ~longword1)
143 | ((longword2 - repeated_one) & ~longword2))
144 & (repeated_one << 7)) != 0)
147 n -= sizeof (longword);
150 char_ptr = (const unsigned char *) longword_ptr;
152 /* At this point, we know that either n < sizeof (longword), or one of the
153 sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On
154 little-endian machines, we could determine the first such byte without
155 any further memory accesses, just by looking at the (tmp1 | tmp2) result
156 from the last loop iteration. But this does not work on big-endian
157 machines. Choose code that works in both cases. */
159 for (; n > 0; --n, ++char_ptr)
161 if (*char_ptr == c1 || *char_ptr == c2)
162 return (void *) char_ptr;