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1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2010
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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 by Roland McGrath (roland@ai.mit.edu).
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9
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10 NOTE: The canonical source of this file is maintained with the GNU C Library.
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11 Bugs can be reported to bug-glibc@prep.ai.mit.edu.
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12
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13 This program is free software: you can redistribute it and/or modify it
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14 under the terms of the GNU General Public License as published by the
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15 Free Software Foundation; either version 3 of the License, or any
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16 later version.
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17
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18 This program is distributed in the hope that it will be useful,
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19 but WITHOUT ANY WARRANTY; without even the implied warranty of
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20 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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21 GNU General Public License for more details.
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22
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23 You should have received a copy of the GNU General Public License
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24 along with this program. If not, see <http://www.gnu.org/licenses/>. */
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25
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26 #ifndef _LIBC
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27 # include <config.h>
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28 #endif
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29
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30 #include <string.h>
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31
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32 #include <stddef.h>
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33
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34 #if defined _LIBC
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35 # include <memcopy.h>
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36 #else
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37 # define reg_char char
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38 #endif
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39
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40 #include <limits.h>
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41
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42 #if HAVE_BP_SYM_H || defined _LIBC
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43 # include <bp-sym.h>
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44 #else
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45 # define BP_SYM(sym) sym
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46 #endif
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47
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48 #undef __memchr
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49 #ifdef _LIBC
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50 # undef memchr
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51 #endif
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52
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53 #ifndef weak_alias
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54 # define __memchr memchr
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55 #endif
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56
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57 /* Search no more than N bytes of S for C. */
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58 void *
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59 __memchr (void const *s, int c_in, size_t n)
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60 {
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61 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
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62 long instead of a 64-bit uintmax_t tends to give better
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63 performance. On 64-bit hardware, unsigned long is generally 64
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64 bits already. Change this typedef to experiment with
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65 performance. */
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66 typedef unsigned long int longword;
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67
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68 const unsigned char *char_ptr;
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69 const longword *longword_ptr;
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70 longword repeated_one;
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71 longword repeated_c;
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72 unsigned reg_char c;
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73
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74 c = (unsigned char) c_in;
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75
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76 /* Handle the first few bytes by reading one byte at a time.
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77 Do this until CHAR_PTR is aligned on a longword boundary. */
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78 for (char_ptr = (const unsigned char *) s;
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79 n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
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80 --n, ++char_ptr)
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81 if (*char_ptr == c)
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82 return (void *) char_ptr;
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83
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84 longword_ptr = (const longword *) char_ptr;
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85
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86 /* All these elucidatory comments refer to 4-byte longwords,
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87 but the theory applies equally well to any size longwords. */
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88
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89 /* Compute auxiliary longword values:
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90 repeated_one is a value which has a 1 in every byte.
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91 repeated_c has c in every byte. */
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92 repeated_one = 0x01010101;
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93 repeated_c = c | (c << 8);
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94 repeated_c |= repeated_c << 16;
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95 if (0xffffffffU < (longword) -1)
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96 {
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97 repeated_one |= repeated_one << 31 << 1;
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98 repeated_c |= repeated_c << 31 << 1;
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99 if (8 < sizeof (longword))
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100 {
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101 size_t i;
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102
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103 for (i = 64; i < sizeof (longword) * 8; i *= 2)
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104 {
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105 repeated_one |= repeated_one << i;
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106 repeated_c |= repeated_c << i;
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107 }
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108 }
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109 }
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110
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111 /* Instead of the traditional loop which tests each byte, we will test a
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112 longword at a time. The tricky part is testing if *any of the four*
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113 bytes in the longword in question are equal to c. We first use an xor
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114 with repeated_c. This reduces the task to testing whether *any of the
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115 four* bytes in longword1 is zero.
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116
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117 We compute tmp =
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118 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
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119 That is, we perform the following operations:
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120 1. Subtract repeated_one.
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121 2. & ~longword1.
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122 3. & a mask consisting of 0x80 in every byte.
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123 Consider what happens in each byte:
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124 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
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125 and step 3 transforms it into 0x80. A carry can also be propagated
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126 to more significant bytes.
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127 - If a byte of longword1 is nonzero, let its lowest 1 bit be at
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128 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1,
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129 the byte ends in a single bit of value 0 and k bits of value 1.
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130 After step 2, the result is just k bits of value 1: 2^k - 1. After
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131 step 3, the result is 0. And no carry is produced.
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132 So, if longword1 has only non-zero bytes, tmp is zero.
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133 Whereas if longword1 has a zero byte, call j the position of the least
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134 significant zero byte. Then the result has a zero at positions 0, ...,
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135 j-1 and a 0x80 at position j. We cannot predict the result at the more
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136 significant bytes (positions j+1..3), but it does not matter since we
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137 already have a non-zero bit at position 8*j+7.
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138
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139 So, the test whether any byte in longword1 is zero is equivalent to
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140 testing whether tmp is nonzero. */
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141
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142 while (n >= sizeof (longword))
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143 {
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144 longword longword1 = *longword_ptr ^ repeated_c;
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145
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146 if ((((longword1 - repeated_one) & ~longword1)
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147 & (repeated_one << 7)) != 0)
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148 break;
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149 longword_ptr++;
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150 n -= sizeof (longword);
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151 }
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152
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153 char_ptr = (const unsigned char *) longword_ptr;
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154
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155 /* At this point, we know that either n < sizeof (longword), or one of the
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156 sizeof (longword) bytes starting at char_ptr is == c. On little-endian
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157 machines, we could determine the first such byte without any further
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158 memory accesses, just by looking at the tmp result from the last loop
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159 iteration. But this does not work on big-endian machines. Choose code
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160 that works in both cases. */
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161
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162 for (; n > 0; --n, ++char_ptr)
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163 {
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164 if (*char_ptr == c)
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165 return (void *) char_ptr;
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166 }
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167
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168 return NULL;
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169 }
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170 #ifdef weak_alias
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171 weak_alias (__memchr, BP_SYM (memchr))
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172 #endif
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