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annotate lib/memchr2.c @ 14610:b427a1938336

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use _GL_ATTRIBUTE_CONST and _GL_ATTRIBUTE_PURE
author Jim Meyering <meyering@redhat.com>
date Sun, 24 Apr 2011 19:02:10 +0200
parents 97fc9a21a8fb
children 6ef4f1f39105
Ignore whitespace changes - Everywhere: Within whitespace: At end of lines:
rev   line source
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1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2011
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2 Free Software Foundation, Inc.
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3
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4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
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5 with help from Dan Sahlin (dan@sics.se) and
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6 commentary by Jim Blandy (jimb@ai.mit.edu);
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7 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
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8 and implemented in glibc by Roland McGrath (roland@ai.mit.edu).
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9 Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
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10
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11 This program is free software: you can redistribute it and/or modify it
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12 under the terms of the GNU General Public License as published by the
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13 Free Software Foundation; either version 3 of the License, or any
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14 later version.
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15
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16 This program is distributed in the hope that it will be useful,
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17 but WITHOUT ANY WARRANTY; without even the implied warranty of
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18 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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19 GNU General Public License for more details.
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20
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21 You should have received a copy of the GNU General Public License
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22 along with this program. If not, see <http://www.gnu.org/licenses/>. */
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23
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24 #include <config.h>
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25
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26 #include "memchr2.h"
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27
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28 #include <limits.h>
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29 #include <stdint.h>
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30 #include <string.h>
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31
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32 /* The attribute __pure__ was added in gcc 2.96. */
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33 #undef _GL_ATTRIBUTE_PURE
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34 #if __GNUC__ > 2 || (__GNUC__ == 2 && __GNUC_MINOR__ >= 96)
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35 # define _GL_ATTRIBUTE_PURE __attribute__ ((__pure__))
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36 #else
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37 # define _GL_ATTRIBUTE_PURE /* empty */
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38 #endif
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39
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40 /* Return the first address of either C1 or C2 (treated as unsigned
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41 char) that occurs within N bytes of the memory region S. If
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42 neither byte appears, return NULL. */
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43 void * _GL_ATTRIBUTE_PURE
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44 memchr2 (void const *s, int c1_in, int c2_in, size_t n)
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45 {
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46 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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47 long instead of a 64-bit uintmax_t tends to give better
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48 performance. On 64-bit hardware, unsigned long is generally 64
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49 bits already. Change this typedef to experiment with
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50 performance. */
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51 typedef unsigned long int longword;
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52
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53 const unsigned char *char_ptr;
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54 const longword *longword_ptr;
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55 longword repeated_one;
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56 longword repeated_c1;
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57 longword repeated_c2;
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58 unsigned char c1;
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59 unsigned char c2;
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60
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61 c1 = (unsigned char) c1_in;
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62 c2 = (unsigned char) c2_in;
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63
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64 if (c1 == c2)
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65 return memchr (s, c1, n);
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66
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67 /* Handle the first few bytes by reading one byte at a time.
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68 Do this until CHAR_PTR is aligned on a longword boundary. */
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69 for (char_ptr = (const unsigned char *) s;
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70 n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
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71 --n, ++char_ptr)
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72 if (*char_ptr == c1 || *char_ptr == c2)
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73 return (void *) char_ptr;
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74
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75 longword_ptr = (const longword *) char_ptr;
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76
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77 /* All these elucidatory comments refer to 4-byte longwords,
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78 but the theory applies equally well to any size longwords. */
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79
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80 /* Compute auxiliary longword values:
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81 repeated_one is a value which has a 1 in every byte.
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82 repeated_c1 has c1 in every byte.
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83 repeated_c2 has c2 in every byte. */
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84 repeated_one = 0x01010101;
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85 repeated_c1 = c1 | (c1 << 8);
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86 repeated_c2 = c2 | (c2 << 8);
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87 repeated_c1 |= repeated_c1 << 16;
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88 repeated_c2 |= repeated_c2 << 16;
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89 if (0xffffffffU < (longword) -1)
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90 {
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91 repeated_one |= repeated_one << 31 << 1;
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92 repeated_c1 |= repeated_c1 << 31 << 1;
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93 repeated_c2 |= repeated_c2 << 31 << 1;
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94 if (8 < sizeof (longword))
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95 {
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96 size_t i;
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97
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98 for (i = 64; i < sizeof (longword) * 8; i *= 2)
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99 {
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100 repeated_one |= repeated_one << i;
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101 repeated_c1 |= repeated_c1 << i;
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102 repeated_c2 |= repeated_c2 << i;
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103 }
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104 }
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105 }
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106
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107 /* Instead of the traditional loop which tests each byte, we will test a
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108 longword at a time. The tricky part is testing if *any of the four*
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109 bytes in the longword in question are equal to c1 or c2. We first use
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110 an xor with repeated_c1 and repeated_c2, respectively. This reduces
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111 the task to testing whether *any of the four* bytes in longword1 or
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112 longword2 is zero.
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113
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114 Let's consider longword1. We compute tmp1 =
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115 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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116 That is, we perform the following operations:
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117 1. Subtract repeated_one.
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118 2. & ~longword1.
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119 3. & a mask consisting of 0x80 in every byte.
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120 Consider what happens in each byte:
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121 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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122 and step 3 transforms it into 0x80. A carry can also be propagated
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123 to more significant bytes.
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124 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
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125 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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126 the byte ends in a single bit of value 0 and k bits of value 1.
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127 After step 2, the result is just k bits of value 1: 2^k - 1. After
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128 step 3, the result is 0. And no carry is produced.
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129 So, if longword1 has only non-zero bytes, tmp1 is zero.
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130 Whereas if longword1 has a zero byte, call j the position of the least
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131 significant zero byte. Then the result has a zero at positions 0, ...,
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132 j-1 and a 0x80 at position j. We cannot predict the result at the more
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133 significant bytes (positions j+1..3), but it does not matter since we
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134 already have a non-zero bit at position 8*j+7.
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135
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136 Similary, we compute tmp2 =
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137 ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
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138
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139 The test whether any byte in longword1 or longword2 is zero is equivalent
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140 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine
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141 this into a single test, whether (tmp1 | tmp2) is nonzero. */
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142
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143 while (n >= sizeof (longword))
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144 {
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145 longword longword1 = *longword_ptr ^ repeated_c1;
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146 longword longword2 = *longword_ptr ^ repeated_c2;
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147
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148 if (((((longword1 - repeated_one) & ~longword1)
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149 | ((longword2 - repeated_one) & ~longword2))
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150 & (repeated_one << 7)) != 0)
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151 break;
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152 longword_ptr++;
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153 n -= sizeof (longword);
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154 }
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155
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156 char_ptr = (const unsigned char *) longword_ptr;
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157
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158 /* At this point, we know that either n < sizeof (longword), or one of the
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159 sizeof (longword) bytes starting at char_ptr is == c1 or == c2. On
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160 little-endian machines, we could determine the first such byte without
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161 any further memory accesses, just by looking at the (tmp1 | tmp2) result
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162 from the last loop iteration. But this does not work on big-endian
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163 machines. Choose code that works in both cases. */
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164
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165 for (; n > 0; --n, ++char_ptr)
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166 {
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167 if (*char_ptr == c1 || *char_ptr == c2)
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168 return (void *) char_ptr;
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169 }
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170
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171 return NULL;
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172 }