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434
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1 /* Searching in a string.
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2 Copyright (C) 2003, 2007, 2008, 2009, 2010 Free Software Foundation, Inc.
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3
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4 This program is free software: you can redistribute it and/or modify
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5 it under the terms of the GNU General Public License as published by
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6 the Free Software Foundation; either version 3 of the License, or
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7 (at your option) any later version.
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8
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9 This program is distributed in the hope that it will be useful,
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10 but WITHOUT ANY WARRANTY; without even the implied warranty of
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11 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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12 GNU General Public License for more details.
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13
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14 You should have received a copy of the GNU General Public License
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15 along with this program. If not, see <http://www.gnu.org/licenses/>. */
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16
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17 #include <config.h>
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18
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19 /* Specification. */
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20 #include <string.h>
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21
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22 /* Find the first occurrence of C in S or the final NUL byte. */
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23 char *
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24 strchrnul (const char *s, int c_in)
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25 {
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26 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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27 long instead of a 64-bit uintmax_t tends to give better
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28 performance. On 64-bit hardware, unsigned long is generally 64
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29 bits already. Change this typedef to experiment with
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30 performance. */
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31 typedef unsigned long int longword;
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32
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33 const unsigned char *char_ptr;
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34 const longword *longword_ptr;
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35 longword repeated_one;
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36 longword repeated_c;
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37 unsigned char c;
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38
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39 c = (unsigned char) c_in;
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40 if (!c)
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41 return rawmemchr (s, 0);
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42
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43 /* Handle the first few bytes by reading one byte at a time.
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44 Do this until CHAR_PTR is aligned on a longword boundary. */
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45 for (char_ptr = (const unsigned char *) s;
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46 (size_t) char_ptr % sizeof (longword) != 0;
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47 ++char_ptr)
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48 if (!*char_ptr || *char_ptr == c)
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49 return (char *) char_ptr;
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50
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51 longword_ptr = (const longword *) char_ptr;
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52
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53 /* All these elucidatory comments refer to 4-byte longwords,
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54 but the theory applies equally well to any size longwords. */
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55
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56 /* Compute auxiliary longword values:
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57 repeated_one is a value which has a 1 in every byte.
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58 repeated_c has c in every byte. */
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59 repeated_one = 0x01010101;
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60 repeated_c = c | (c << 8);
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61 repeated_c |= repeated_c << 16;
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62 if (0xffffffffU < (longword) -1)
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63 {
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64 repeated_one |= repeated_one << 31 << 1;
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65 repeated_c |= repeated_c << 31 << 1;
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66 if (8 < sizeof (longword))
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67 {
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68 size_t i;
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69
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70 for (i = 64; i < sizeof (longword) * 8; i *= 2)
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71 {
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72 repeated_one |= repeated_one << i;
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73 repeated_c |= repeated_c << i;
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74 }
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75 }
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76 }
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77
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78 /* Instead of the traditional loop which tests each byte, we will
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79 test a longword at a time. The tricky part is testing if *any of
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80 the four* bytes in the longword in question are equal to NUL or
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81 c. We first use an xor with repeated_c. This reduces the task
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82 to testing whether *any of the four* bytes in longword1 or
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83 longword2 is zero.
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84
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85 Let's consider longword1. We compute tmp =
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86 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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87 That is, we perform the following operations:
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88 1. Subtract repeated_one.
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89 2. & ~longword1.
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90 3. & a mask consisting of 0x80 in every byte.
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91 Consider what happens in each byte:
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92 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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93 and step 3 transforms it into 0x80. A carry can also be propagated
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94 to more significant bytes.
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95 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
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96 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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97 the byte ends in a single bit of value 0 and k bits of value 1.
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98 After step 2, the result is just k bits of value 1: 2^k - 1. After
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99 step 3, the result is 0. And no carry is produced.
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100 So, if longword1 has only non-zero bytes, tmp is zero.
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101 Whereas if longword1 has a zero byte, call j the position of the least
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102 significant zero byte. Then the result has a zero at positions 0, ...,
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103 j-1 and a 0x80 at position j. We cannot predict the result at the more
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104 significant bytes (positions j+1..3), but it does not matter since we
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105 already have a non-zero bit at position 8*j+7.
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106
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107 The test whether any byte in longword1 or longword2 is zero is equivalent
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108 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
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109 this into a single test, whether (tmp1 | tmp2) is nonzero.
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110
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111 This test can read more than one byte beyond the end of a string,
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112 depending on where the terminating NUL is encountered. However,
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113 this is considered safe since the initialization phase ensured
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114 that the read will be aligned, therefore, the read will not cross
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115 page boundaries and will not cause a fault. */
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116
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117 while (1)
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118 {
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119 longword longword1 = *longword_ptr ^ repeated_c;
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120 longword longword2 = *longword_ptr;
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121
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122 if (((((longword1 - repeated_one) & ~longword1)
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123 | ((longword2 - repeated_one) & ~longword2))
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124 & (repeated_one << 7)) != 0)
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125 break;
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126 longword_ptr++;
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127 }
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128
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129 char_ptr = (const unsigned char *) longword_ptr;
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130
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131 /* At this point, we know that one of the sizeof (longword) bytes
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132 starting at char_ptr is == 0 or == c. On little-endian machines,
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133 we could determine the first such byte without any further memory
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134 accesses, just by looking at the tmp result from the last loop
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135 iteration. But this does not work on big-endian machines.
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136 Choose code that works in both cases. */
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137
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138 char_ptr = (unsigned char *) longword_ptr;
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139 while (*char_ptr && (*char_ptr != c))
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140 char_ptr++;
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141 return (char *) char_ptr;
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142 }
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