1 /* memrchr -- find the last occurrence of a byte in a memory block
2
3 Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2018 Free Software
4 Foundation, Inc.
5
6 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
7 with help from Dan Sahlin (dan@sics.se) and
8 commentary by Jim Blandy (jimb@ai.mit.edu);
9 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
10 and implemented by Roland McGrath (roland@ai.mit.edu).
11
12 This program is free software: you can redistribute it and/or modify
13 it under the terms of the GNU General Public License as published by
14 the Free Software Foundation; either version 3 of the License, or
15 (at your option) any later version.
16
17 This program is distributed in the hope that it will be useful,
18 but WITHOUT ANY WARRANTY; without even the implied warranty of
19 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
20 GNU General Public License for more details.
21
22 You should have received a copy of the GNU General Public License
23 along with this program. If not, see <https://www.gnu.org/licenses/>. */
24
25 #if defined _LIBC
26 # include <memcopy.h>
27 #else
28 # include <config.h>
29 # define reg_char char
30 #endif
31
32 #include <string.h>
33 #include <limits.h>
34
35 #undef __memrchr
36 #ifdef _LIBC
37 # undef memrchr
38 #endif
39
40 #ifndef weak_alias
41 # define __memrchr memrchr
42 #endif
43
44 /* Search no more than N bytes of S for C. */
45 void *
__memrchr(void const * s,int c_in,size_t n)46 __memrchr (void const *s, int c_in, size_t n)
47 {
48 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
49 long instead of a 64-bit uintmax_t tends to give better
50 performance. On 64-bit hardware, unsigned long is generally 64
51 bits already. Change this typedef to experiment with
52 performance. */
53 typedef unsigned long int longword;
54
55 const unsigned char *char_ptr;
56 const longword *longword_ptr;
57 longword repeated_one;
58 longword repeated_c;
59 unsigned reg_char c;
60
61 c = (unsigned char) c_in;
62
63 /* Handle the last few bytes by reading one byte at a time.
64 Do this until CHAR_PTR is aligned on a longword boundary. */
65 for (char_ptr = (const unsigned char *) s + n;
66 n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
67 --n)
68 if (*--char_ptr == c)
69 return (void *) char_ptr;
70
71 longword_ptr = (const longword *) char_ptr;
72
73 /* All these elucidatory comments refer to 4-byte longwords,
74 but the theory applies equally well to any size longwords. */
75
76 /* Compute auxiliary longword values:
77 repeated_one is a value which has a 1 in every byte.
78 repeated_c has c in every byte. */
79 repeated_one = 0x01010101;
80 repeated_c = c | (c << 8);
81 repeated_c |= repeated_c << 16;
82 if (0xffffffffU < (longword) -1)
83 {
84 repeated_one |= repeated_one << 31 << 1;
85 repeated_c |= repeated_c << 31 << 1;
86 if (8 < sizeof (longword))
87 {
88 size_t i;
89
90 for (i = 64; i < sizeof (longword) * 8; i *= 2)
91 {
92 repeated_one |= repeated_one << i;
93 repeated_c |= repeated_c << i;
94 }
95 }
96 }
97
98 /* Instead of the traditional loop which tests each byte, we will test a
99 longword at a time. The tricky part is testing if *any of the four*
100 bytes in the longword in question are equal to c. We first use an xor
101 with repeated_c. This reduces the task to testing whether *any of the
102 four* bytes in longword1 is zero.
103
104 We compute tmp =
105 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
106 That is, we perform the following operations:
107 1. Subtract repeated_one.
108 2. & ~longword1.
109 3. & a mask consisting of 0x80 in every byte.
110 Consider what happens in each byte:
111 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
112 and step 3 transforms it into 0x80. A carry can also be propagated
113 to more significant bytes.
114 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
115 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
116 the byte ends in a single bit of value 0 and k bits of value 1.
117 After step 2, the result is just k bits of value 1: 2^k - 1. After
118 step 3, the result is 0. And no carry is produced.
119 So, if longword1 has only non-zero bytes, tmp is zero.
120 Whereas if longword1 has a zero byte, call j the position of the least
121 significant zero byte. Then the result has a zero at positions 0, ...,
122 j-1 and a 0x80 at position j. We cannot predict the result at the more
123 significant bytes (positions j+1..3), but it does not matter since we
124 already have a non-zero bit at position 8*j+7.
125
126 So, the test whether any byte in longword1 is zero is equivalent to
127 testing whether tmp is nonzero. */
128
129 while (n >= sizeof (longword))
130 {
131 longword longword1 = *--longword_ptr ^ repeated_c;
132
133 if ((((longword1 - repeated_one) & ~longword1)
134 & (repeated_one << 7)) != 0)
135 {
136 longword_ptr++;
137 break;
138 }
139 n -= sizeof (longword);
140 }
141
142 char_ptr = (const unsigned char *) longword_ptr;
143
144 /* At this point, we know that either n < sizeof (longword), or one of the
145 sizeof (longword) bytes starting at char_ptr is == c. On little-endian
146 machines, we could determine the first such byte without any further
147 memory accesses, just by looking at the tmp result from the last loop
148 iteration. But this does not work on big-endian machines. Choose code
149 that works in both cases. */
150
151 while (n-- > 0)
152 {
153 if (*--char_ptr == c)
154 return (void *) char_ptr;
155 }
156
157 return NULL;
158 }
159 #ifdef weak_alias
160 weak_alias (__memrchr, memrchr)
161 #endif
162