1/* Copyright (C) 1991 Free Software Foundation, Inc. 2 Based on strlen implemention by Torbjorn Granlund (tege@sics.se), 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 8The GNU C Library is free software; you can redistribute it and/or 9modify it under the terms of the GNU Library General Public License as 10published by the Free Software Foundation; either version 2 of the 11License, or (at your option) any later version. 12 13The GNU C Library is distributed in the hope that it will be useful, 14but WITHOUT ANY WARRANTY; without even the implied warranty of 15MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU 16Library General Public License for more details. 17 18You should have received a copy of the GNU Library General Public 19License along with the GNU C Library; see the file COPYING.LIB. If 20not, write to the Free Software Foundation, Inc., 59 Temple Place - 21Suite 330, Boston, MA 02111-1307, USA. 22 23This file was modified slightly by Ian Lance Taylor, May 1992, for 24Taylor UUCP. It assumes 32 bit longs. I'm willing to trust that any 25system which does not have 32 bit longs will have its own 26implementation of memchr. */ 27 28#include "uucp.h" 29 30/* Search no more than N bytes of S for C. */ 31 32pointer 33memchr (s, c, n) 34 constpointer s; 35 int c; 36 size_t n; 37{ 38 const char *char_ptr; 39 const unsigned long int *longword_ptr; 40 unsigned long int longword, magic_bits, charmask; 41 42 c = BUCHAR (c); 43 44 /* Handle the first few characters by reading one character at a time. 45 Do this until CHAR_PTR is aligned on a 4-byte border. */ 46 for (char_ptr = s; n > 0 && ((unsigned long int) char_ptr & 3) != 0; 47 --n, ++char_ptr) 48 if (BUCHAR (*char_ptr) == c) 49 return (pointer) char_ptr; 50 51 longword_ptr = (unsigned long int *) char_ptr; 52 53 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits 54 the "holes." Note that there is a hole just to the left of 55 each byte, with an extra at the end: 56 57 bits: 01111110 11111110 11111110 11111111 58 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD 59 60 The 1-bits make sure that carries propagate to the next 0-bit. 61 The 0-bits provide holes for carries to fall into. */ 62 magic_bits = 0x7efefeff; 63 64 /* Set up a longword, each of whose bytes is C. */ 65 charmask = c | (c << 8); 66 charmask |= charmask << 16; 67 68 /* Instead of the traditional loop which tests each character, 69 we will test a longword at a time. The tricky part is testing 70 if *any of the four* bytes in the longword in question are zero. */ 71 while (n >= 4) 72 { 73 /* We tentatively exit the loop if adding MAGIC_BITS to 74 LONGWORD fails to change any of the hole bits of LONGWORD. 75 76 1) Is this safe? Will it catch all the zero bytes? 77 Suppose there is a byte with all zeros. Any carry bits 78 propagating from its left will fall into the hole at its 79 least significant bit and stop. Since there will be no 80 carry from its most significant bit, the LSB of the 81 byte to the left will be unchanged, and the zero will be 82 detected. 83 84 2) Is this worthwhile? Will it ignore everything except 85 zero bytes? Suppose every byte of LONGWORD has a bit set 86 somewhere. There will be a carry into bit 8. If bit 8 87 is set, this will carry into bit 16. If bit 8 is clear, 88 one of bits 9-15 must be set, so there will be a carry 89 into bit 16. Similarly, there will be a carry into bit 90 24. If one of bits 24-30 is set, there will be a carry 91 into bit 31, so all of the hole bits will be changed. 92 93 The one misfire occurs when bits 24-30 are clear and bit 94 31 is set; in this case, the hole at bit 31 is not 95 changed. If we had access to the processor carry flag, 96 we could close this loophole by putting the fourth hole 97 at bit 32! 98 99 So it ignores everything except 128's, when they're aligned 100 properly. 101 102 3) But wait! Aren't we looking for C, not zero? 103 Good point. So what we do is XOR LONGWORD with a longword, 104 each of whose bytes is C. This turns each byte that is C 105 into a zero. */ 106 107 longword = *longword_ptr++ ^ charmask; 108 109 /* Add MAGIC_BITS to LONGWORD. */ 110 if ((((longword + magic_bits) 111 112 /* Set those bits that were unchanged by the addition. */ 113 ^ ~longword) 114 115 /* Look at only the hole bits. If any of the hole bits 116 are unchanged, most likely one of the bytes was a 117 zero. */ 118 & ~magic_bits) != 0) 119 { 120 /* Which of the bytes was C? If none of them were, it was 121 a misfire; continue the search. */ 122 123 const char *cp = (const char *) (longword_ptr - 1); 124 125 if (BUCHAR (cp[0]) == c) 126 return (pointer) cp; 127 if (BUCHAR (cp[1]) == c) 128 return (pointer) &cp[1]; 129 if (BUCHAR (cp[2]) == c) 130 return (pointer) &cp[2]; 131 if (BUCHAR (cp[3]) == c) 132 return (pointer) &cp[3]; 133 } 134 135 n -= 4; 136 } 137 138 char_ptr = (const char *) longword_ptr; 139 140 while (n-- > 0) 141 { 142 if (BUCHAR (*char_ptr) == c) 143 return (pointer) char_ptr; 144 else 145 ++char_ptr; 146 } 147 148 return NULL; 149} 150