Mercurial > hg > octave-kai > gnulib-hg
annotate lib/memchr2.c @ 17476:6057744acd2c default tip master
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author | Karl Berry <karl@freefriends.org> |
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date | Fri, 16 Aug 2013 06:32:22 -0700 (2013-08-16) |
parents | 2760c772c42c |
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1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2013 |
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2 Free Software Foundation, Inc. |
9742 | 3 |
4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se), | |
5 with help from Dan Sahlin (dan@sics.se) and | |
6 commentary by Jim Blandy (jimb@ai.mit.edu); | |
7 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu), | |
8 and implemented in glibc by Roland McGrath (roland@ai.mit.edu). | |
9 Extension to memchr2 implemented by Eric Blake (ebb9@byu.net). | |
10 | |
11 This program is free software: you can redistribute it and/or modify it | |
12 under the terms of the GNU General Public License as published by the | |
13 Free Software Foundation; either version 3 of the License, or any | |
14 later version. | |
15 | |
16 This program is distributed in the hope that it will be useful, | |
17 but WITHOUT ANY WARRANTY; without even the implied warranty of | |
18 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the | |
19 GNU General Public License for more details. | |
20 | |
21 You should have received a copy of the GNU General Public License | |
22 along with this program. If not, see <http://www.gnu.org/licenses/>. */ | |
23 | |
24 #include <config.h> | |
25 | |
26 #include "memchr2.h" | |
27 | |
28 #include <limits.h> | |
29 #include <stdint.h> | |
30 #include <string.h> | |
31 | |
32 /* Return the first address of either C1 or C2 (treated as unsigned | |
33 char) that occurs within N bytes of the memory region S. If | |
34 neither byte appears, return NULL. */ | |
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35 void * |
9742 | 36 memchr2 (void const *s, int c1_in, int c2_in, size_t n) |
37 { | |
9967 | 38 /* On 32-bit hardware, choosing longword to be a 32-bit unsigned |
39 long instead of a 64-bit uintmax_t tends to give better | |
40 performance. On 64-bit hardware, unsigned long is generally 64 | |
41 bits already. Change this typedef to experiment with | |
42 performance. */ | |
9995 | 43 typedef unsigned long int longword; |
9967 | 44 |
9742 | 45 const unsigned char *char_ptr; |
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46 void const *void_ptr; |
9967 | 47 const longword *longword_ptr; |
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48 longword repeated_one; |
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49 longword repeated_c1; |
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50 longword repeated_c2; |
9742 | 51 unsigned char c1; |
52 unsigned char c2; | |
53 | |
54 c1 = (unsigned char) c1_in; | |
55 c2 = (unsigned char) c2_in; | |
56 | |
57 if (c1 == c2) | |
58 return memchr (s, c1, n); | |
59 | |
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60 /* Handle the first few bytes by reading one byte at a time. |
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61 Do this until VOID_PTR is aligned on a longword boundary. */ |
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62 for (void_ptr = s; |
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63 n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0; |
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64 --n) |
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65 { |
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66 char_ptr = void_ptr; |
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67 if (*char_ptr == c1 || *char_ptr == c2) |
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68 return (void *) void_ptr; |
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69 void_ptr = char_ptr + 1; |
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70 } |
9742 | 71 |
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72 longword_ptr = void_ptr; |
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73 |
9742 | 74 /* All these elucidatory comments refer to 4-byte longwords, |
75 but the theory applies equally well to any size longwords. */ | |
76 | |
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77 /* Compute auxiliary longword values: |
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78 repeated_one is a value which has a 1 in every byte. |
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79 repeated_c1 has c1 in every byte. |
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80 repeated_c2 has c2 in every byte. */ |
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81 repeated_one = 0x01010101; |
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82 repeated_c1 = c1 | (c1 << 8); |
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83 repeated_c2 = c2 | (c2 << 8); |
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84 repeated_c1 |= repeated_c1 << 16; |
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85 repeated_c2 |= repeated_c2 << 16; |
9995 | 86 if (0xffffffffU < (longword) -1) |
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87 { |
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88 repeated_one |= repeated_one << 31 << 1; |
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89 repeated_c1 |= repeated_c1 << 31 << 1; |
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90 repeated_c2 |= repeated_c2 << 31 << 1; |
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91 if (8 < sizeof (longword)) |
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92 { |
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93 size_t i; |
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94 |
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95 for (i = 64; i < sizeof (longword) * 8; i *= 2) |
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96 { |
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97 repeated_one |= repeated_one << i; |
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98 repeated_c1 |= repeated_c1 << i; |
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99 repeated_c2 |= repeated_c2 << i; |
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100 } |
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101 } |
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102 } |
9742 | 103 |
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104 /* Instead of the traditional loop which tests each byte, we will test a |
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105 longword at a time. The tricky part is testing if *any of the four* |
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106 bytes in the longword in question are equal to c1 or c2. We first use |
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107 an xor with repeated_c1 and repeated_c2, respectively. This reduces |
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108 the task to testing whether *any of the four* bytes in longword1 or |
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109 longword2 is zero. |
9967 | 110 |
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111 Let's consider longword1. We compute tmp1 = |
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112 ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). |
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113 That is, we perform the following operations: |
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114 1. Subtract repeated_one. |
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115 2. & ~longword1. |
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116 3. & a mask consisting of 0x80 in every byte. |
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117 Consider what happens in each byte: |
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118 - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, |
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119 and step 3 transforms it into 0x80. A carry can also be propagated |
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120 to more significant bytes. |
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121 - If a byte of longword1 is nonzero, let its lowest 1 bit be at |
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122 position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, |
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123 the byte ends in a single bit of value 0 and k bits of value 1. |
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124 After step 2, the result is just k bits of value 1: 2^k - 1. After |
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125 step 3, the result is 0. And no carry is produced. |
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126 So, if longword1 has only non-zero bytes, tmp1 is zero. |
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127 Whereas if longword1 has a zero byte, call j the position of the least |
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128 significant zero byte. Then the result has a zero at positions 0, ..., |
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129 j-1 and a 0x80 at position j. We cannot predict the result at the more |
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130 significant bytes (positions j+1..3), but it does not matter since we |
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131 already have a non-zero bit at position 8*j+7. |
9967 | 132 |
16358 | 133 Similarly, we compute tmp2 = |
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134 ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7). |
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135 |
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136 The test whether any byte in longword1 or longword2 is zero is equivalent |
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137 to testing whether tmp1 is nonzero or tmp2 is nonzero. We can combine |
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138 this into a single test, whether (tmp1 | tmp2) is nonzero. */ |
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139 |
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140 while (n >= sizeof (longword)) |
9742 | 141 { |
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142 longword longword1 = *longword_ptr ^ repeated_c1; |
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143 longword longword2 = *longword_ptr ^ repeated_c2; |
9742 | 144 |
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145 if (((((longword1 - repeated_one) & ~longword1) |
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146 | ((longword2 - repeated_one) & ~longword2)) |
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147 & (repeated_one << 7)) != 0) |
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148 break; |
9967 | 149 longword_ptr++; |
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150 n -= sizeof (longword); |
9742 | 151 } |
152 | |
153 char_ptr = (const unsigned char *) longword_ptr; | |
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 == c1 or == c2. On |
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157 little-endian machines, we could determine the first such byte without |
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158 any further memory accesses, just by looking at the (tmp1 | tmp2) result |
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159 from the last loop iteration. But this does not work on big-endian |
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160 machines. Choose code that works in both cases. */ |
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161 |
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162 for (; n > 0; --n, ++char_ptr) |
9742 | 163 { |
164 if (*char_ptr == c1 || *char_ptr == c2) | |
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165 return (void *) char_ptr; |
9742 | 166 } |
167 | |
9967 | 168 return NULL; |
9742 | 169 } |