1 /* Searching in a string. 2 Copyright (C) 2003, 2007-2012 Free Software Foundation, Inc. 3 4 This program is free software: you can redistribute it and/or modify 5 it under the terms of the GNU General Public License as published by 6 the Free Software Foundation; either version 3 of the License, or 7 (at your option) any later version. 8 9 This program 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 General Public License for more details. 13 14 You should have received a copy of the GNU General Public License 15 along with this program. If not, see <http://www.gnu.org/licenses/>. */ 16 17 #include <config.h> 18 19 /* Specification. */ 20 #include <string.h> 21 22 /* Find the first occurrence of C in S or the final NUL byte. */ 23 char * 24 strchrnul (const char *s, int c_in) 25 { 26 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned 27 long instead of a 64-bit uintmax_t tends to give better 28 performance. On 64-bit hardware, unsigned long is generally 64 29 bits already. Change this typedef to experiment with 30 performance. */ 31 typedef unsigned long int longword; 32 33 const unsigned char *char_ptr; 34 const longword *longword_ptr; 35 longword repeated_one; 36 longword repeated_c; 37 unsigned char c; 38 39 c = (unsigned char) c_in; 40 if (!c) 41 return rawmemchr (s, 0); 42 43 /* Handle the first few bytes by reading one byte at a time. 44 Do this until CHAR_PTR is aligned on a longword boundary. */ 45 for (char_ptr = (const unsigned char *) s; 46 (size_t) char_ptr % sizeof (longword) != 0; 47 ++char_ptr) 48 if (!*char_ptr || *char_ptr == c) 49 return (char *) char_ptr; 50 51 longword_ptr = (const longword *) char_ptr; 52 53 /* All these elucidatory comments refer to 4-byte longwords, 54 but the theory applies equally well to any size longwords. */ 55 56 /* Compute auxiliary longword values: 57 repeated_one is a value which has a 1 in every byte. 58 repeated_c has c in every byte. */ 59 repeated_one = 0x01010101; 60 repeated_c = c | (c << 8); 61 repeated_c |= repeated_c << 16; 62 if (0xffffffffU < (longword) -1) 63 { 64 repeated_one |= repeated_one << 31 << 1; 65 repeated_c |= repeated_c << 31 << 1; 66 if (8 < sizeof (longword)) 67 { 68 size_t i; 69 70 for (i = 64; i < sizeof (longword) * 8; i *= 2) 71 { 72 repeated_one |= repeated_one << i; 73 repeated_c |= repeated_c << i; 74 } 75 } 76 } 77 78 /* Instead of the traditional loop which tests each byte, we will 79 test a longword at a time. The tricky part is testing if *any of 80 the four* bytes in the longword in question are equal to NUL or 81 c. We first use an xor with repeated_c. This reduces the task 82 to testing whether *any of the four* bytes in longword1 or 83 longword2 is zero. 84 85 Let's consider longword1. We compute tmp = 86 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). 87 That is, we perform the following operations: 88 1. Subtract repeated_one. 89 2. & ~longword1. 90 3. & a mask consisting of 0x80 in every byte. 91 Consider what happens in each byte: 92 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, 93 and step 3 transforms it into 0x80. A carry can also be propagated 94 to more significant bytes. 95 - If a byte of longword1 is nonzero, let its lowest 1 bit be at 96 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, 97 the byte ends in a single bit of value 0 and k bits of value 1. 98 After step 2, the result is just k bits of value 1: 2^k - 1. After 99 step 3, the result is 0. And no carry is produced. 100 So, if longword1 has only non-zero bytes, tmp is zero. 101 Whereas if longword1 has a zero byte, call j the position of the least 102 significant zero byte. Then the result has a zero at positions 0, ..., 103 j-1 and a 0x80 at position j. We cannot predict the result at the more 104 significant bytes (positions j+1..3), but it does not matter since we 105 already have a non-zero bit at position 8*j+7. 106 107 The test whether any byte in longword1 or longword2 is zero is equivalent 108 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine 109 this into a single test, whether (tmp1 | tmp2) is nonzero. 110 111 This test can read more than one byte beyond the end of a string, 112 depending on where the terminating NUL is encountered. However, 113 this is considered safe since the initialization phase ensured 114 that the read will be aligned, therefore, the read will not cross 115 page boundaries and will not cause a fault. */ 116 117 while (1) 118 { 119 longword longword1 = *longword_ptr ^ repeated_c; 120 longword longword2 = *longword_ptr; 121 122 if (((((longword1 - repeated_one) & ~longword1) 123 | ((longword2 - repeated_one) & ~longword2)) 124 & (repeated_one << 7)) != 0) 125 break; 126 longword_ptr++; 127 } 128 129 char_ptr = (const unsigned char *) longword_ptr; 130 131 /* At this point, we know that one of the sizeof (longword) bytes 132 starting at char_ptr is == 0 or == c. On little-endian machines, 133 we could determine the first such byte without any further memory 134 accesses, just by looking at the tmp result from the last loop 135 iteration. But this does not work on big-endian machines. 136 Choose code that works in both cases. */ 137 138 char_ptr = (unsigned char *) longword_ptr; 139 while (*char_ptr && (*char_ptr != c)) 140 char_ptr++; 141 return (char *) char_ptr; 142 } 143
