1 /* Copyright (C) 1991, 1993 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).
8 The GNU C Library is free software; you can redistribute it and/or
9 modify it under the terms of the GNU Library General Public License as
10 published by the Free Software Foundation; either version 2 of the
11 License, or (at your option) any later version.
13 The GNU C Library is distributed in the hope that it will be useful,
14 but WITHOUT ANY WARRANTY; without even the implied warranty of
15 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
16 Library General Public License for more details.
18 You should have received a copy of the GNU Library General Public
19 License along with the GNU C Library; see the file COPYING.LIB. If
20 not, write to the Free Software Foundation, Inc., 675 Mass Ave,
21 Cambridge, MA 02139, USA. */
27 #if (SIZEOF_LONG != 4 && SIZEOF_LONG != 8)
28 error This function works only on systems for which sizeof(long) is 4 or 8.
29 /* The previous line would begin with `#error,' but some compilers can't
30 handle that even when the condition is false. */
33 /* Search no more than N bytes of S for C. */
41 const unsigned char *char_ptr;
42 const unsigned long int *longword_ptr;
43 unsigned long int longword, magic_bits, charmask;
45 c = (unsigned char) c;
47 /* Handle the first few characters by reading one character at a time.
48 Do this until CHAR_PTR is aligned on a longword boundary. */
49 for (char_ptr = s; n > 0 && ((unsigned long int) char_ptr
50 & (sizeof (longword) - 1)) != 0;
53 return (char *) char_ptr;
55 /* All these elucidatory comments refer to 4-byte longwords,
56 but the theory applies equally well to 8-byte longwords. */
58 longword_ptr = (unsigned long int *) char_ptr;
60 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
61 the "holes." Note that there is a hole just to the left of
62 each byte, with an extra at the end:
64 bits: 01111110 11111110 11111110 11111111
65 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
67 The 1-bits make sure that carries propagate to the next 0-bit.
68 The 0-bits provide holes for carries to fall into. */
69 #if (SIZEOF_LONG == 8)
70 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
72 magic_bits = 0x7efefeff;
73 #endif /* SIZEOF_LONG == 8 */
75 /* Set up a longword, each of whose bytes is C. */
76 charmask = c | (c << 8);
77 charmask |= charmask << 16;
78 #if (SIZEOF_LONG == 8)
79 charmask |= charmask << 32;
80 #endif /* SIZEOF_LONG == 8 */
81 if (sizeof (longword) > 8)
85 /* Instead of the traditional loop which tests each character,
86 we will test a longword at a time. The tricky part is testing
87 if *any of the four* bytes in the longword in question are zero. */
88 while (n >= sizeof (longword))
90 /* We tentatively exit the loop if adding MAGIC_BITS to
91 LONGWORD fails to change any of the hole bits of LONGWORD.
93 1) Is this safe? Will it catch all the zero bytes?
94 Suppose there is a byte with all zeros. Any carry bits
95 propagating from its left will fall into the hole at its
96 least significant bit and stop. Since there will be no
97 carry from its most significant bit, the LSB of the
98 byte to the left will be unchanged, and the zero will be
101 2) Is this worthwhile? Will it ignore everything except
102 zero bytes? Suppose every byte of LONGWORD has a bit set
103 somewhere. There will be a carry into bit 8. If bit 8
104 is set, this will carry into bit 16. If bit 8 is clear,
105 one of bits 9-15 must be set, so there will be a carry
106 into bit 16. Similarly, there will be a carry into bit
107 24. If one of bits 24-30 is set, there will be a carry
108 into bit 31, so all of the hole bits will be changed.
110 The one misfire occurs when bits 24-30 are clear and bit
111 31 is set; in this case, the hole at bit 31 is not
112 changed. If we had access to the processor carry flag,
113 we could close this loophole by putting the fourth hole
116 So it ignores everything except 128's, when they're aligned
119 3) But wait! Aren't we looking for C, not zero?
120 Good point. So what we do is XOR LONGWORD with a longword,
121 each of whose bytes is C. This turns each byte that is C
124 longword = *longword_ptr++ ^ charmask;
126 /* Add MAGIC_BITS to LONGWORD. */
127 if ((((longword + magic_bits)
129 /* Set those bits that were unchanged by the addition. */
132 /* Look at only the hole bits. If any of the hole bits
133 are unchanged, most likely one of the bytes was a
137 /* Which of the bytes was C? If none of them were, it was
138 a misfire; continue the search. */
140 const unsigned char *cp = (const unsigned char *) (longword_ptr - 1);
145 return (char *) &cp[1];
147 return (char *) &cp[2];
149 return (char *) &cp[3];
150 #if (SIZEOF_LONG == 8)
152 return (char *) &cp[4];
154 return (char *) &cp[5];
156 return (char *) &cp[6];
158 return (char *) &cp[7];
159 #endif /* SIZEOF_LONG == 8 */
162 n -= sizeof (longword);
165 char_ptr = (const unsigned char *) longword_ptr;
170 return (char *) char_ptr;