Mercurial > hg > octave-kai > gnulib-hg
annotate lib/memchr.c @ 4664:42147e1c0cee
Remove K&R cruft.
| author | Paul Eggert <eggert@cs.ucla.edu> |
|---|---|
| date | Tue, 09 Sep 2003 22:22:46 +0000 |
| parents | e422838277f4 |
| children | a535859efd14 |
| rev | line source |
|---|---|
| 4664 | 1 /* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003 Free |
| 2 Software Foundation, Inc. | |
| 3 | |
| 884 | 4 Based on strlen implementation by Torbjorn Granlund (tege@sics.se), |
| 14 | 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 by Roland McGrath (roland@ai.mit.edu). | |
| 9 | |
| 394 | 10 NOTE: The canonical source of this file is maintained with the GNU C Library. |
| 11 Bugs can be reported to bug-glibc@prep.ai.mit.edu. | |
| 14 | 12 |
| 394 | 13 This program is free software; you can redistribute it and/or modify it |
| 14 under the terms of the GNU General Public License as published by the | |
| 15 Free Software Foundation; either version 2, or (at your option) any | |
| 16 later version. | |
| 17 | |
| 18 This program is distributed in the hope that it will be useful, | |
| 14 | 19 but WITHOUT ANY WARRANTY; without even the implied warranty of |
| 394 | 20 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the |
| 21 GNU General Public License for more details. | |
| 14 | 22 |
| 394 | 23 You should have received a copy of the GNU General Public License |
| 24 along with this program; if not, write to the Free Software | |
|
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25 Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, |
|
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26 USA. */ |
| 14 | 27 |
| 166 | 28 #ifdef HAVE_CONFIG_H |
| 1872 | 29 # include <config.h> |
| 166 | 30 #endif |
| 31 | |
| 4664 | 32 #include <string.h> |
| 394 | 33 |
| 2932 | 34 #if defined _LIBC |
| 35 # include <memcopy.h> | |
| 36 #else | |
| 37 # define reg_char char | |
| 448 | 38 #endif |
| 39 | |
| 4664 | 40 #include <limits.h> |
| 41 #include <stdlib.h> | |
| 394 | 42 |
| 43 #define LONG_MAX_32_BITS 2147483647 | |
| 44 | |
| 45 #include <sys/types.h> | |
| 2932 | 46 #if HAVE_BP_SYM_H || defined _LIBC |
| 47 # include <bp-sym.h> | |
| 48 #else | |
| 49 # define BP_SYM(sym) sym | |
| 50 #endif | |
| 394 | 51 |
| 2932 | 52 #undef memchr |
| 53 #undef __memchr | |
| 394 | 54 |
| 14 | 55 /* Search no more than N bytes of S for C. */ |
| 4664 | 56 void * |
| 57 __memchr (void const *s, int c_in, size_t n) | |
| 14 | 58 { |
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59 const unsigned char *char_ptr; |
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60 const unsigned long int *longword_ptr; |
| 14 | 61 unsigned long int longword, magic_bits, charmask; |
| 2932 | 62 unsigned reg_char c; |
| 14 | 63 |
| 2932 | 64 c = (unsigned char) c_in; |
| 14 | 65 |
| 66 /* Handle the first few characters by reading one character at a time. | |
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67 Do this until CHAR_PTR is aligned on a longword boundary. */ |
| 448 | 68 for (char_ptr = (const unsigned char *) s; |
| 69 n > 0 && ((unsigned long int) char_ptr | |
| 70 & (sizeof (longword) - 1)) != 0; | |
| 14 | 71 --n, ++char_ptr) |
| 72 if (*char_ptr == c) | |
| 4664 | 73 return (void *) char_ptr; |
| 14 | 74 |
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75 /* All these elucidatory comments refer to 4-byte longwords, |
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76 but the theory applies equally well to 8-byte longwords. */ |
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77 |
| 14 | 78 longword_ptr = (unsigned long int *) char_ptr; |
| 79 | |
| 80 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits | |
| 81 the "holes." Note that there is a hole just to the left of | |
| 82 each byte, with an extra at the end: | |
| 448 | 83 |
| 14 | 84 bits: 01111110 11111110 11111110 11111111 |
| 448 | 85 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD |
| 14 | 86 |
| 87 The 1-bits make sure that carries propagate to the next 0-bit. | |
| 88 The 0-bits provide holes for carries to fall into. */ | |
| 394 | 89 |
| 90 if (sizeof (longword) != 4 && sizeof (longword) != 8) | |
| 91 abort (); | |
| 92 | |
| 93 #if LONG_MAX <= LONG_MAX_32_BITS | |
| 94 magic_bits = 0x7efefeff; | |
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95 #else |
| 394 | 96 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff; |
| 97 #endif | |
| 14 | 98 |
| 99 /* Set up a longword, each of whose bytes is C. */ | |
| 100 charmask = c | (c << 8); | |
| 101 charmask |= charmask << 16; | |
| 394 | 102 #if LONG_MAX > LONG_MAX_32_BITS |
| 400 | 103 charmask |= charmask << 32; |
| 394 | 104 #endif |
| 14 | 105 |
| 106 /* Instead of the traditional loop which tests each character, | |
| 107 we will test a longword at a time. The tricky part is testing | |
| 108 if *any of the four* bytes in the longword in question are zero. */ | |
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109 while (n >= sizeof (longword)) |
| 14 | 110 { |
| 111 /* We tentatively exit the loop if adding MAGIC_BITS to | |
| 112 LONGWORD fails to change any of the hole bits of LONGWORD. | |
| 113 | |
| 114 1) Is this safe? Will it catch all the zero bytes? | |
| 115 Suppose there is a byte with all zeros. Any carry bits | |
| 116 propagating from its left will fall into the hole at its | |
| 117 least significant bit and stop. Since there will be no | |
| 118 carry from its most significant bit, the LSB of the | |
| 119 byte to the left will be unchanged, and the zero will be | |
| 120 detected. | |
| 121 | |
| 122 2) Is this worthwhile? Will it ignore everything except | |
| 123 zero bytes? Suppose every byte of LONGWORD has a bit set | |
| 124 somewhere. There will be a carry into bit 8. If bit 8 | |
| 125 is set, this will carry into bit 16. If bit 8 is clear, | |
| 126 one of bits 9-15 must be set, so there will be a carry | |
| 127 into bit 16. Similarly, there will be a carry into bit | |
| 128 24. If one of bits 24-30 is set, there will be a carry | |
| 129 into bit 31, so all of the hole bits will be changed. | |
| 130 | |
| 131 The one misfire occurs when bits 24-30 are clear and bit | |
| 132 31 is set; in this case, the hole at bit 31 is not | |
| 133 changed. If we had access to the processor carry flag, | |
| 134 we could close this loophole by putting the fourth hole | |
| 135 at bit 32! | |
| 136 | |
| 137 So it ignores everything except 128's, when they're aligned | |
| 138 properly. | |
| 139 | |
| 140 3) But wait! Aren't we looking for C, not zero? | |
| 141 Good point. So what we do is XOR LONGWORD with a longword, | |
| 142 each of whose bytes is C. This turns each byte that is C | |
| 143 into a zero. */ | |
| 144 | |
| 145 longword = *longword_ptr++ ^ charmask; | |
| 146 | |
| 147 /* Add MAGIC_BITS to LONGWORD. */ | |
| 148 if ((((longword + magic_bits) | |
| 448 | 149 |
| 394 | 150 /* Set those bits that were unchanged by the addition. */ |
| 14 | 151 ^ ~longword) |
| 448 | 152 |
| 394 | 153 /* Look at only the hole bits. If any of the hole bits |
| 14 | 154 are unchanged, most likely one of the bytes was a |
| 155 zero. */ | |
| 156 & ~magic_bits) != 0) | |
| 157 { | |
| 158 /* Which of the bytes was C? If none of them were, it was | |
| 159 a misfire; continue the search. */ | |
| 160 | |
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161 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1); |
| 14 | 162 |
| 163 if (cp[0] == c) | |
| 4664 | 164 return (void *) cp; |
| 14 | 165 if (cp[1] == c) |
| 4664 | 166 return (void *) &cp[1]; |
| 14 | 167 if (cp[2] == c) |
| 4664 | 168 return (void *) &cp[2]; |
| 14 | 169 if (cp[3] == c) |
| 4664 | 170 return (void *) &cp[3]; |
| 394 | 171 #if LONG_MAX > 2147483647 |
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172 if (cp[4] == c) |
| 4664 | 173 return (void *) &cp[4]; |
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174 if (cp[5] == c) |
| 4664 | 175 return (void *) &cp[5]; |
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176 if (cp[6] == c) |
| 4664 | 177 return (void *) &cp[6]; |
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178 if (cp[7] == c) |
| 4664 | 179 return (void *) &cp[7]; |
| 394 | 180 #endif |
| 14 | 181 } |
| 182 | |
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183 n -= sizeof (longword); |
| 14 | 184 } |
| 185 | |
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186 char_ptr = (const unsigned char *) longword_ptr; |
| 14 | 187 |
| 188 while (n-- > 0) | |
| 189 { | |
| 190 if (*char_ptr == c) | |
| 4664 | 191 return (void *) char_ptr; |
| 14 | 192 else |
| 193 ++char_ptr; | |
| 194 } | |
| 195 | |
| 196 return 0; | |
| 197 } | |
| 2932 | 198 #ifdef weak_alias |
| 199 weak_alias (__memchr, BP_SYM (memchr)) | |
| 200 #endif |