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annotate lib/memchr.c @ 2807:807294ed0f4f
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| author | Jim Meyering <jim@meyering.net> |
|---|---|
| date | Mon, 07 Aug 2000 15:48:18 +0000 |
| parents | 5994c6f939c5 |
| children | e422838277f4 |
| rev | line source |
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1 /* Copyright (C) 1991, 1993, 1997, 1999 Free Software Foundation, Inc. |
| 884 | 2 Based on strlen implementation by Torbjorn Granlund (tege@sics.se), |
| 14 | 3 with help from Dan Sahlin (dan@sics.se) and |
| 4 commentary by Jim Blandy (jimb@ai.mit.edu); | |
| 5 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu), | |
| 6 and implemented by Roland McGrath (roland@ai.mit.edu). | |
| 7 | |
| 394 | 8 NOTE: The canonical source of this file is maintained with the GNU C Library. |
| 9 Bugs can be reported to bug-glibc@prep.ai.mit.edu. | |
| 14 | 10 |
| 394 | 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 2, or (at your option) any | |
| 14 later version. | |
| 15 | |
| 16 This program is distributed in the hope that it will be useful, | |
| 14 | 17 but WITHOUT ANY WARRANTY; without even the implied warranty of |
| 394 | 18 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the |
| 19 GNU General Public License for more details. | |
| 14 | 20 |
| 394 | 21 You should have received a copy of the GNU General Public License |
| 22 along with this program; if not, write to the Free Software | |
|
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23 Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, |
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24 USA. */ |
| 14 | 25 |
| 166 | 26 #ifdef HAVE_CONFIG_H |
| 1872 | 27 # include <config.h> |
| 166 | 28 #endif |
| 29 | |
| 394 | 30 #undef __ptr_t |
| 31 #if defined (__cplusplus) || (defined (__STDC__) && __STDC__) | |
| 32 # define __ptr_t void * | |
| 33 #else /* Not C++ or ANSI C. */ | |
| 34 # define __ptr_t char * | |
| 35 #endif /* C++ or ANSI C. */ | |
| 36 | |
| 448 | 37 #if defined (_LIBC) |
| 38 # include <string.h> | |
| 39 #endif | |
| 40 | |
| 443 | 41 #if defined (HAVE_LIMITS_H) || defined (_LIBC) |
| 42 # include <limits.h> | |
| 394 | 43 #endif |
| 44 | |
| 45 #define LONG_MAX_32_BITS 2147483647 | |
| 46 | |
| 47 #ifndef LONG_MAX | |
| 1872 | 48 # define LONG_MAX LONG_MAX_32_BITS |
| 394 | 49 #endif |
| 50 | |
| 51 #include <sys/types.h> | |
| 52 | |
| 53 | |
| 14 | 54 /* Search no more than N bytes of S for C. */ |
| 55 | |
| 394 | 56 __ptr_t |
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57 memchr (s, c, n) |
| 394 | 58 const __ptr_t s; |
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59 int c; |
| 394 | 60 size_t n; |
| 14 | 61 { |
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62 const unsigned char *char_ptr; |
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63 const unsigned long int *longword_ptr; |
| 14 | 64 unsigned long int longword, magic_bits, charmask; |
| 65 | |
| 66 c = (unsigned char) c; | |
| 67 | |
| 68 /* Handle the first few characters by reading one character at a time. | |
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69 Do this until CHAR_PTR is aligned on a longword boundary. */ |
| 448 | 70 for (char_ptr = (const unsigned char *) s; |
| 71 n > 0 && ((unsigned long int) char_ptr | |
| 72 & (sizeof (longword) - 1)) != 0; | |
| 14 | 73 --n, ++char_ptr) |
| 74 if (*char_ptr == c) | |
| 394 | 75 return (__ptr_t) char_ptr; |
| 14 | 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 8-byte longwords. */ |
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79 |
| 14 | 80 longword_ptr = (unsigned long int *) char_ptr; |
| 81 | |
| 82 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits | |
| 83 the "holes." Note that there is a hole just to the left of | |
| 84 each byte, with an extra at the end: | |
| 448 | 85 |
| 14 | 86 bits: 01111110 11111110 11111110 11111111 |
| 448 | 87 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD |
| 14 | 88 |
| 89 The 1-bits make sure that carries propagate to the next 0-bit. | |
| 90 The 0-bits provide holes for carries to fall into. */ | |
| 394 | 91 |
| 92 if (sizeof (longword) != 4 && sizeof (longword) != 8) | |
| 93 abort (); | |
| 94 | |
| 95 #if LONG_MAX <= LONG_MAX_32_BITS | |
| 96 magic_bits = 0x7efefeff; | |
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97 #else |
| 394 | 98 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff; |
| 99 #endif | |
| 14 | 100 |
| 101 /* Set up a longword, each of whose bytes is C. */ | |
| 102 charmask = c | (c << 8); | |
| 103 charmask |= charmask << 16; | |
| 394 | 104 #if LONG_MAX > LONG_MAX_32_BITS |
| 400 | 105 charmask |= charmask << 32; |
| 394 | 106 #endif |
| 14 | 107 |
| 108 /* Instead of the traditional loop which tests each character, | |
| 109 we will test a longword at a time. The tricky part is testing | |
| 110 if *any of the four* bytes in the longword in question are zero. */ | |
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111 while (n >= sizeof (longword)) |
| 14 | 112 { |
| 113 /* We tentatively exit the loop if adding MAGIC_BITS to | |
| 114 LONGWORD fails to change any of the hole bits of LONGWORD. | |
| 115 | |
| 116 1) Is this safe? Will it catch all the zero bytes? | |
| 117 Suppose there is a byte with all zeros. Any carry bits | |
| 118 propagating from its left will fall into the hole at its | |
| 119 least significant bit and stop. Since there will be no | |
| 120 carry from its most significant bit, the LSB of the | |
| 121 byte to the left will be unchanged, and the zero will be | |
| 122 detected. | |
| 123 | |
| 124 2) Is this worthwhile? Will it ignore everything except | |
| 125 zero bytes? Suppose every byte of LONGWORD has a bit set | |
| 126 somewhere. There will be a carry into bit 8. If bit 8 | |
| 127 is set, this will carry into bit 16. If bit 8 is clear, | |
| 128 one of bits 9-15 must be set, so there will be a carry | |
| 129 into bit 16. Similarly, there will be a carry into bit | |
| 130 24. If one of bits 24-30 is set, there will be a carry | |
| 131 into bit 31, so all of the hole bits will be changed. | |
| 132 | |
| 133 The one misfire occurs when bits 24-30 are clear and bit | |
| 134 31 is set; in this case, the hole at bit 31 is not | |
| 135 changed. If we had access to the processor carry flag, | |
| 136 we could close this loophole by putting the fourth hole | |
| 137 at bit 32! | |
| 138 | |
| 139 So it ignores everything except 128's, when they're aligned | |
| 140 properly. | |
| 141 | |
| 142 3) But wait! Aren't we looking for C, not zero? | |
| 143 Good point. So what we do is XOR LONGWORD with a longword, | |
| 144 each of whose bytes is C. This turns each byte that is C | |
| 145 into a zero. */ | |
| 146 | |
| 147 longword = *longword_ptr++ ^ charmask; | |
| 148 | |
| 149 /* Add MAGIC_BITS to LONGWORD. */ | |
| 150 if ((((longword + magic_bits) | |
| 448 | 151 |
| 394 | 152 /* Set those bits that were unchanged by the addition. */ |
| 14 | 153 ^ ~longword) |
| 448 | 154 |
| 394 | 155 /* Look at only the hole bits. If any of the hole bits |
| 14 | 156 are unchanged, most likely one of the bytes was a |
| 157 zero. */ | |
| 158 & ~magic_bits) != 0) | |
| 159 { | |
| 160 /* Which of the bytes was C? If none of them were, it was | |
| 161 a misfire; continue the search. */ | |
| 162 | |
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163 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1); |
| 14 | 164 |
| 165 if (cp[0] == c) | |
| 394 | 166 return (__ptr_t) cp; |
| 14 | 167 if (cp[1] == c) |
| 394 | 168 return (__ptr_t) &cp[1]; |
| 14 | 169 if (cp[2] == c) |
| 394 | 170 return (__ptr_t) &cp[2]; |
| 14 | 171 if (cp[3] == c) |
| 394 | 172 return (__ptr_t) &cp[3]; |
| 173 #if LONG_MAX > 2147483647 | |
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174 if (cp[4] == c) |
| 394 | 175 return (__ptr_t) &cp[4]; |
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176 if (cp[5] == c) |
| 394 | 177 return (__ptr_t) &cp[5]; |
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178 if (cp[6] == c) |
| 394 | 179 return (__ptr_t) &cp[6]; |
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180 if (cp[7] == c) |
| 394 | 181 return (__ptr_t) &cp[7]; |
| 182 #endif | |
| 14 | 183 } |
| 184 | |
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185 n -= sizeof (longword); |
| 14 | 186 } |
| 187 | |
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188 char_ptr = (const unsigned char *) longword_ptr; |
| 14 | 189 |
| 190 while (n-- > 0) | |
| 191 { | |
| 192 if (*char_ptr == c) | |
| 394 | 193 return (__ptr_t) char_ptr; |
| 14 | 194 else |
| 195 ++char_ptr; | |
| 196 } | |
| 197 | |
| 198 return 0; | |
| 199 } |