1 /* Searching in a string.
2 Copyright (C) 2008-2023 Free Software Foundation, Inc.
3
4 This file is free software: you can redistribute it and/or modify
5 it under the terms of the GNU Lesser General Public License as
6 published by the Free Software Foundation; either version 2.1 of the
7 License, or (at your option) any later version.
8
9 This file is distributed in the hope that it will be useful,
10 but WITHOUT ANY WARRANTY; without even the implied warranty of
11 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
12 GNU Lesser General Public License for more details.
13
14 You should have received a copy of the GNU Lesser General Public License
15 along with this program. If not, see <https://www.gnu.org/licenses/>. */
16
17 #include <config.h>
18
19 /* Specification. */
20 #include <string.h>
21
22 /* A function definition is needed only if HAVE_RAWMEMCHR is not defined. */
23 #if !HAVE_RAWMEMCHR
24
25 # include <limits.h>
26 # include <stdint.h>
27
28
29 /* Find the first occurrence of C in S. */
30 void *
31 rawmemchr (const void *s, int c_in)
32 {
33 # ifdef __CHERI_PURE_CAPABILITY__
34 /* Most architectures let you read an aligned word,
35 even if the unsigned char array at S ends in the middle of the word.
36 However CHERI does not, so call memchr
37 with the underlying object's remaining length.
38 This cannot return NULL if S points to a C_IN-terminated array.
39 Use builtins rather than including <cheri.h> which is less stable. */
40 return memchr (s, c_in, (__builtin_cheri_length_get (s)
41 - __builtin_cheri_offset_get (s)));
42 # else
43
44 /* You can change this typedef to experiment with performance. */
45 typedef uintptr_t longword;
46 /* Verify that the longword type lacks padding bits. */
47 static_assert (UINTPTR_WIDTH == UCHAR_WIDTH * sizeof (uintptr_t));
48
49 const unsigned char *char_ptr;
50 unsigned char c = c_in;
51
52 /* Handle the first few bytes by reading one byte at a time.
53 Do this until CHAR_PTR is aligned on a natural longword boundary,
54 as using alignof (longword) might be slower. */
55 for (char_ptr = (const unsigned char *) s;
56 (uintptr_t) char_ptr % sizeof (longword) != 0;
57 ++char_ptr)
58 if (*char_ptr == c)
59 return (void *) char_ptr;
60
61 longword const *longword_ptr = s = char_ptr;
62
63 /* Compute auxiliary longword values:
64 repeated_one is a value which has a 1 in every byte.
65 repeated_c has c in every byte. */
66 longword repeated_one = (longword) -1 / UCHAR_MAX;
67 longword repeated_c = repeated_one * c;
68 longword repeated_hibit = repeated_one * (UCHAR_MAX / 2 + 1);
69
70 /* Instead of the traditional loop which tests each byte, we will
71 test a longword at a time. The tricky part is testing if any of
72 the bytes in the longword in question are equal to
73 c. We first use an xor with repeated_c. This reduces the task
74 to testing whether any of the bytes in longword1 is zero.
75
76 (The following comments assume 8-bit bytes, as POSIX requires;
77 the code's use of UCHAR_MAX should work even if bytes have more
78 than 8 bits.)
79
80 We compute tmp =
81 ((longword1 - repeated_one) & ~longword1) & (repeated_one * 0x80).
82 That is, we perform the following operations:
83 1. Subtract repeated_one.
84 2. & ~longword1.
85 3. & a mask consisting of 0x80 in every byte.
86 Consider what happens in each byte:
87 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
88 and step 3 transforms it into 0x80. A carry can also be propagated
89 to more significant bytes.
90 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
91 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
92 the byte ends in a single bit of value 0 and k bits of value 1.
93 After step 2, the result is just k bits of value 1: 2^k - 1. After
94 step 3, the result is 0. And no carry is produced.
95 So, if longword1 has only non-zero bytes, tmp is zero.
96 Whereas if longword1 has a zero byte, call j the position of the least
97 significant zero byte. Then the result has a zero at positions 0, ...,
98 j-1 and a 0x80 at position j. We cannot predict the result at the more
99 significant bytes (positions j+1..3), but it does not matter since we
100 already have a non-zero bit at position 8*j+7.
101
102 The test whether any byte in longword1 is zero is equivalent
103 to testing whether tmp is nonzero.
104
105 This test can read beyond the end of a string, depending on where
106 C_IN is encountered. However, this is considered safe since the
107 initialization phase ensured that the read will be aligned,
108 therefore, the read will not cross page boundaries and will not
109 cause a fault. */
110
111 while (1)
112 {
113 longword longword1 = *longword_ptr ^ repeated_c;
114
115 if ((((longword1 - repeated_one) & ~longword1) & repeated_hibit) != 0)
116 break;
117 longword_ptr++;
118 }
119
120 char_ptr = s = longword_ptr;
121
122 /* At this point, we know that one of the sizeof (longword) bytes
123 starting at char_ptr is == c. If we knew endianness, we
124 could determine the first such byte without any further memory
125 accesses, just by looking at the tmp result from the last loop
126 iteration. However, the following simple and portable code does
127 not attempt this potential optimization. */
128
129 while (*char_ptr != c)
130 char_ptr++;
131 return (void *) char_ptr;
132 # endif
133 }
134
135 #endif