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annotate lib/rawmemchr.c @ 16358:a712776b11ce

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maint: spelling fixes
author Paul Eggert <eggert@cs.ucla.edu>
date Sun, 05 Feb 2012 13:42:03 -0800
parents 8250f2777afc
children e542fd46ad6f
Ignore whitespace changes - Everywhere: Within whitespace: At end of lines:
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1 /* Searching in a string.
16201
8250f2777afc maint: update all copyright year number ranges
Jim Meyering <meyering@redhat.com>
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2 Copyright (C) 2008-2012 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. */
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23 void *
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24 rawmemchr (const void *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
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41 /* Handle the first few bytes by reading one byte at a time.
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42 Do this until CHAR_PTR is aligned on a longword boundary. */
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43 for (char_ptr = (const unsigned char *) s;
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44 (size_t) char_ptr % sizeof (longword) != 0;
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45 ++char_ptr)
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46 if (*char_ptr == c)
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47 return (void *) char_ptr;
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48
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49 longword_ptr = (const longword *) char_ptr;
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50
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51 /* All these elucidatory comments refer to 4-byte longwords,
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52 but the theory applies equally well to any size longwords. */
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53
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54 /* Compute auxiliary longword values:
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55 repeated_one is a value which has a 1 in every byte.
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56 repeated_c has c in every byte. */
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57 repeated_one = 0x01010101;
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58 repeated_c = c | (c << 8);
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59 repeated_c |= repeated_c << 16;
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60 if (0xffffffffU < (longword) -1)
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61 {
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62 repeated_one |= repeated_one << 31 << 1;
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63 repeated_c |= repeated_c << 31 << 1;
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64 if (8 < sizeof (longword))
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65 {
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66 size_t i;
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67
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68 for (i = 64; i < sizeof (longword) * 8; i *= 2)
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69 {
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70 repeated_one |= repeated_one << i;
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71 repeated_c |= repeated_c << i;
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72 }
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73 }
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74 }
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75
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76 /* Instead of the traditional loop which tests each byte, we will
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77 test a longword at a time. The tricky part is testing if *any of
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78 the four* bytes in the longword in question are equal to NUL or
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79 c. We first use an xor with repeated_c. This reduces the task
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80 to testing whether *any of the four* bytes in longword1 is zero.
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81
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82 We compute tmp =
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83 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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84 That is, we perform the following operations:
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85 1. Subtract repeated_one.
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86 2. & ~longword1.
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87 3. & a mask consisting of 0x80 in every byte.
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88 Consider what happens in each byte:
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89 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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90 and step 3 transforms it into 0x80. A carry can also be propagated
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91 to more significant bytes.
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92 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
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93 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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94 the byte ends in a single bit of value 0 and k bits of value 1.
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95 After step 2, the result is just k bits of value 1: 2^k - 1. After
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96 step 3, the result is 0. And no carry is produced.
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97 So, if longword1 has only non-zero bytes, tmp is zero.
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98 Whereas if longword1 has a zero byte, call j the position of the least
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99 significant zero byte. Then the result has a zero at positions 0, ...,
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100 j-1 and a 0x80 at position j. We cannot predict the result at the more
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101 significant bytes (positions j+1..3), but it does not matter since we
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102 already have a non-zero bit at position 8*j+7.
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103
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104 The test whether any byte in longword1 is zero is equivalent
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105 to testing whether tmp is nonzero.
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106
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107 This test can read beyond the end of a string, depending on where
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108 C_IN is encountered. However, this is considered safe since the
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109 initialization phase ensured that the read will be aligned,
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110 therefore, the read will not cross page boundaries and will not
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111 cause a fault. */
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112
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113 while (1)
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114 {
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115 longword longword1 = *longword_ptr ^ repeated_c;
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116
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117 if ((((longword1 - repeated_one) & ~longword1)
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118 & (repeated_one << 7)) != 0)
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119 break;
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120 longword_ptr++;
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121 }
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122
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123 char_ptr = (const unsigned char *) longword_ptr;
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124
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125 /* At this point, we know that one of the sizeof (longword) bytes
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126 starting at char_ptr is == c. On little-endian machines, we
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127 could determine the first such byte without any further memory
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128 accesses, just by looking at the tmp result from the last loop
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129 iteration. But this does not work on big-endian machines.
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130 Choose code that works in both cases. */
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131
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132 char_ptr = (unsigned char *) longword_ptr;
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133 while (*char_ptr != c)
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134 char_ptr++;
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135 return (void *) char_ptr;
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136 }