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, see
22 <http://www.gnu.org/licenses/>. */
32 #define LONG_MAX_32_BITS 2147483647
34 /* Search no more than N bytes of S for C. */
35 void *memrchr (const void * s, int c_in, size_t n)
37 const unsigned char *char_ptr;
38 const unsigned long int *longword_ptr;
39 unsigned long int longword, magic_bits, charmask;
42 c = (unsigned char) c_in;
44 /* Handle the last few characters by reading one character at a time.
45 Do this until CHAR_PTR is aligned on a longword boundary. */
46 for (char_ptr = (const unsigned char *) s + n;
47 n > 0 && ((unsigned long int) char_ptr
48 & (sizeof (longword) - 1)) != 0;
51 return (void *) char_ptr;
53 /* All these elucidatory comments refer to 4-byte longwords,
54 but the theory applies equally well to 8-byte longwords. */
56 longword_ptr = (const unsigned long int *) char_ptr;
58 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
59 the "holes." Note that there is a hole just to the left of
60 each byte, with an extra at the end:
62 bits: 01111110 11111110 11111110 11111111
63 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
65 The 1-bits make sure that carries propagate to the next 0-bit.
66 The 0-bits provide holes for carries to fall into. */
68 if (sizeof (longword) != 4 && sizeof (longword) != 8)
71 #if LONG_MAX <= LONG_MAX_32_BITS
72 magic_bits = 0x7efefeff;
74 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
77 /* Set up a longword, each of whose bytes is C. */
78 charmask = c | (c << 8);
79 charmask |= charmask << 16;
80 #if LONG_MAX > LONG_MAX_32_BITS
81 charmask |= charmask << 32;
84 /* Instead of the traditional loop which tests each character,
85 we will test a longword at a time. The tricky part is testing
86 if *any of the four* bytes in the longword in question are zero. */
87 while (n >= sizeof (longword))
89 /* We tentatively exit the loop if adding MAGIC_BITS to
90 LONGWORD fails to change any of the hole bits of LONGWORD.
92 1) Is this safe? Will it catch all the zero bytes?
93 Suppose there is a byte with all zeros. Any carry bits
94 propagating from its left will fall into the hole at its
95 least significant bit and stop. Since there will be no
96 carry from its most significant bit, the LSB of the
97 byte to the left will be unchanged, and the zero will be
100 2) Is this worthwhile? Will it ignore everything except
101 zero bytes? Suppose every byte of LONGWORD has a bit set
102 somewhere. There will be a carry into bit 8. If bit 8
103 is set, this will carry into bit 16. If bit 8 is clear,
104 one of bits 9-15 must be set, so there will be a carry
105 into bit 16. Similarly, there will be a carry into bit
106 24. If one of bits 24-30 is set, there will be a carry
107 into bit 31, so all of the hole bits will be changed.
109 The one misfire occurs when bits 24-30 are clear and bit
110 31 is set; in this case, the hole at bit 31 is not
111 changed. If we had access to the processor carry flag,
112 we could close this loophole by putting the fourth hole
115 So it ignores everything except 128's, when they're aligned
118 3) But wait! Aren't we looking for C, not zero?
119 Good point. So what we do is XOR LONGWORD with a longword,
120 each of whose bytes is C. This turns each byte that is C
123 longword = *--longword_ptr ^ charmask;
125 /* Add MAGIC_BITS to LONGWORD. */
126 if ((((longword + magic_bits)
128 /* Set those bits that were unchanged by the addition. */
131 /* Look at only the hole bits. If any of the hole bits
132 are unchanged, most likely one of the bytes was a
136 /* Which of the bytes was C? If none of them were, it was
137 a misfire; continue the search. */
139 const unsigned char *cp = (const unsigned char *) longword_ptr;
141 #if LONG_MAX > 2147483647
143 return (void *) &cp[7];
145 return (void *) &cp[6];
147 return (void *) &cp[5];
149 return (void *) &cp[4];
152 return (void *) &cp[3];
154 return (void *) &cp[2];
156 return (void *) &cp[1];
161 n -= sizeof (longword);
164 char_ptr = (const unsigned char *) longword_ptr;
168 if (*--char_ptr == c)
169 return (void *) char_ptr;
174 libc_hidden_def(memrchr)