1 /* memrchr -- find the last occurrence of a byte in a memory block
2 Copyright (C) 1991, 93, 96, 97, 99, 2000 Free Software Foundation, Inc.
3 This file is part of the GNU C Library.
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 by Roland McGrath (roland@ai.mit.edu).
10 The GNU C Library is free software; you can redistribute it and/or
11 modify it under the terms of the GNU Lesser General Public
12 License as published by the Free Software Foundation; either
13 version 2.1 of the License, or (at your option) any later version.
15 The GNU C Library is distributed in the hope that it will be useful,
16 but WITHOUT ANY WARRANTY; without even the implied warranty of
17 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
18 Lesser General Public License for more details.
20 You should have received a copy of the GNU Lesser General Public
21 License along with the GNU C Library; if not, write to the Free
22 Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
29 libc_hidden_proto(memrchr)
30 libc_hidden_proto(abort)
34 #define LONG_MAX_32_BITS 2147483647
36 /* Search no more than N bytes of S for C. */
37 void *memrchr (const void * s, int c_in, size_t n)
39 const unsigned char *char_ptr;
40 const unsigned long int *longword_ptr;
41 unsigned long int longword, magic_bits, charmask;
44 c = (unsigned char) c_in;
46 /* Handle the last few characters by reading one character at a time.
47 Do this until CHAR_PTR is aligned on a longword boundary. */
48 for (char_ptr = (const unsigned char *) s + n;
49 n > 0 && ((unsigned long int) char_ptr
50 & (sizeof (longword) - 1)) != 0;
53 return (void *) 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 = (const 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. */
70 if (sizeof (longword) != 4 && sizeof (longword) != 8)
73 #if LONG_MAX <= LONG_MAX_32_BITS
74 magic_bits = 0x7efefeff;
76 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
79 /* Set up a longword, each of whose bytes is C. */
80 charmask = c | (c << 8);
81 charmask |= charmask << 16;
82 #if LONG_MAX > LONG_MAX_32_BITS
83 charmask |= charmask << 32;
86 /* Instead of the traditional loop which tests each character,
87 we will test a longword at a time. The tricky part is testing
88 if *any of the four* bytes in the longword in question are zero. */
89 while (n >= sizeof (longword))
91 /* We tentatively exit the loop if adding MAGIC_BITS to
92 LONGWORD fails to change any of the hole bits of LONGWORD.
94 1) Is this safe? Will it catch all the zero bytes?
95 Suppose there is a byte with all zeros. Any carry bits
96 propagating from its left will fall into the hole at its
97 least significant bit and stop. Since there will be no
98 carry from its most significant bit, the LSB of the
99 byte to the left will be unchanged, and the zero will be
102 2) Is this worthwhile? Will it ignore everything except
103 zero bytes? Suppose every byte of LONGWORD has a bit set
104 somewhere. There will be a carry into bit 8. If bit 8
105 is set, this will carry into bit 16. If bit 8 is clear,
106 one of bits 9-15 must be set, so there will be a carry
107 into bit 16. Similarly, there will be a carry into bit
108 24. If one of bits 24-30 is set, there will be a carry
109 into bit 31, so all of the hole bits will be changed.
111 The one misfire occurs when bits 24-30 are clear and bit
112 31 is set; in this case, the hole at bit 31 is not
113 changed. If we had access to the processor carry flag,
114 we could close this loophole by putting the fourth hole
117 So it ignores everything except 128's, when they're aligned
120 3) But wait! Aren't we looking for C, not zero?
121 Good point. So what we do is XOR LONGWORD with a longword,
122 each of whose bytes is C. This turns each byte that is C
125 longword = *--longword_ptr ^ charmask;
127 /* Add MAGIC_BITS to LONGWORD. */
128 if ((((longword + magic_bits)
130 /* Set those bits that were unchanged by the addition. */
133 /* Look at only the hole bits. If any of the hole bits
134 are unchanged, most likely one of the bytes was a
138 /* Which of the bytes was C? If none of them were, it was
139 a misfire; continue the search. */
141 const unsigned char *cp = (const unsigned char *) longword_ptr;
143 #if LONG_MAX > 2147483647
145 return (void *) &cp[7];
147 return (void *) &cp[6];
149 return (void *) &cp[5];
151 return (void *) &cp[4];
154 return (void *) &cp[3];
156 return (void *) &cp[2];
158 return (void *) &cp[1];
163 n -= sizeof (longword);
166 char_ptr = (const unsigned char *) longword_ptr;
170 if (*--char_ptr == c)
171 return (void *) char_ptr;
176 libc_hidden_def(memrchr)