1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2021
2 Free Software Foundation, Inc.
3
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).
10
11 This file is free software: you can redistribute it and/or modify
12 it under the terms of the GNU Lesser General Public License as
13 published by the Free Software Foundation; either version 2.1 of the
14 License, or (at your option) any later version.
15
16 This file 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 Lesser General Public License for more details.
20
21 You should have received a copy of the GNU Lesser General Public License
22 along with this program. If not, see <https://www.gnu.org/licenses/>. */
23
24 #include <config.h>
25
26 #include "memchr2.h"
27
28 #include <limits.h>
29 #include <stdint.h>
30 #include <string.h>
31
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. */
35 void *
memchr2(void const * s,int c1_in,int c2_in,size_t n)36 memchr2 (void const *s, int c1_in, int c2_in, size_t n)
37 {
38 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
39 long instead of a 64-bit uintmax_t tends to give better
40 performance. On 64-bit hardware, unsigned long is generally 64
41 bits already. Change this typedef to experiment with
42 performance. */
43 typedef unsigned long int longword;
44
45 const unsigned char *char_ptr;
46 void const *void_ptr;
47 const longword *longword_ptr;
48 longword repeated_one;
49 longword repeated_c1;
50 longword repeated_c2;
51 unsigned char c1;
52 unsigned char c2;
53
54 c1 = (unsigned char) c1_in;
55 c2 = (unsigned char) c2_in;
56
57 if (c1 == c2)
58 return memchr (s, c1, n);
59
60 /* Handle the first few bytes by reading one byte at a time.
61 Do this until VOID_PTR is aligned on a longword boundary. */
62 for (void_ptr = s;
63 n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;
64 --n)
65 {
66 char_ptr = void_ptr;
67 if (*char_ptr == c1 || *char_ptr == c2)
68 return (void *) void_ptr;
69 void_ptr = char_ptr + 1;
70 }
71
72 longword_ptr = void_ptr;
73
74 /* All these elucidatory comments refer to 4-byte longwords,
75 but the theory applies equally well to any size longwords. */
76
77 /* Compute auxiliary longword values:
78 repeated_one is a value which has a 1 in every byte.
79 repeated_c1 has c1 in every byte.
80 repeated_c2 has c2 in every byte. */
81 repeated_one = 0x01010101;
82 repeated_c1 = c1 | (c1 << 8);
83 repeated_c2 = c2 | (c2 << 8);
84 repeated_c1 |= repeated_c1 << 16;
85 repeated_c2 |= repeated_c2 << 16;
86 if (0xffffffffU < (longword) -1)
87 {
88 repeated_one |= repeated_one << 31 << 1;
89 repeated_c1 |= repeated_c1 << 31 << 1;
90 repeated_c2 |= repeated_c2 << 31 << 1;
91 if (8 < sizeof (longword))
92 {
93 size_t i;
94
95 for (i = 64; i < sizeof (longword) * 8; i *= 2)
96 {
97 repeated_one |= repeated_one << i;
98 repeated_c1 |= repeated_c1 << i;
99 repeated_c2 |= repeated_c2 << i;
100 }
101 }
102 }
103
104 /* Instead of the traditional loop which tests each byte, we will test a
105 longword at a time. The tricky part is testing if *any of the four*
106 bytes in the longword in question are equal to c1 or c2. We first use
107 an xor with repeated_c1 and repeated_c2, respectively. This reduces
108 the task to testing whether *any of the four* bytes in longword1 or
109 longword2 is zero.
110
111 Let's consider longword1. We compute tmp1 =
112 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
113 That is, we perform the following operations:
114 1. Subtract repeated_one.
115 2. & ~longword1.
116 3. & a mask consisting of 0x80 in every byte.
117 Consider what happens in each byte:
118 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
119 and step 3 transforms it into 0x80. A carry can also be propagated
120 to more significant bytes.
121 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
122 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
123 the byte ends in a single bit of value 0 and k bits of value 1.
124 After step 2, the result is just k bits of value 1: 2^k - 1. After
125 step 3, the result is 0. And no carry is produced.
126 So, if longword1 has only non-zero bytes, tmp1 is zero.
127 Whereas if longword1 has a zero byte, call j the position of the least
128 significant zero byte. Then the result has a zero at positions 0, ...,
129 j-1 and a 0x80 at position j. We cannot predict the result at the more
130 significant bytes (positions j+1..3), but it does not matter since we
131 already have a non-zero bit at position 8*j+7.
132
133 Similarly, we compute tmp2 =
134 ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
135
136 The test whether any byte in longword1 or longword2 is zero is equivalent
137 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
138 this into a single test, whether (tmp1 | tmp2) is nonzero. */
139
140 while (n >= sizeof (longword))
141 {
142 longword longword1 = *longword_ptr ^ repeated_c1;
143 longword longword2 = *longword_ptr ^ repeated_c2;
144
145 if (((((longword1 - repeated_one) & ~longword1)
146 | ((longword2 - repeated_one) & ~longword2))
147 & (repeated_one << 7)) != 0)
148 break;
149 longword_ptr++;
150 n -= sizeof (longword);
151 }
152
153 char_ptr = (const unsigned char *) longword_ptr;
154
155 /* At this point, we know that either n < sizeof (longword), or one of the
156 sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On
157 little-endian machines, we could determine the first such byte without
158 any further memory accesses, just by looking at the (tmp1 | tmp2) result
159 from the last loop iteration. But this does not work on big-endian
160 machines. Choose code that works in both cases. */
161
162 for (; n > 0; --n, ++char_ptr)
163 {
164 if (*char_ptr == c1 || *char_ptr == c2)
165 return (void *) char_ptr;
166 }
167
168 return NULL;
169 }
170