1 /* memrchr -- find the last occurrence of a byte in a memory block
3 Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005,
4 2006, 2007 Free Software Foundation, Inc.
6 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
7 with help from Dan Sahlin (dan@sics.se) and
8 commentary by Jim Blandy (jimb@ai.mit.edu);
9 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
10 and implemented by Roland McGrath (roland@ai.mit.edu).
12 This program is free software: you can redistribute it and/or modify
13 it under the terms of the GNU General Public License as published by
14 the Free Software Foundation; either version 3 of the License, or
15 (at your option) any later version.
17 This program is distributed in the hope that it will be useful,
18 but WITHOUT ANY WARRANTY; without even the implied warranty of
19 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
20 GNU General Public License for more details.
22 You should have received a copy of the GNU General Public License
23 along with this program. If not, see <http://www.gnu.org/licenses/>. */
29 # define reg_char char
39 # define __memrchr memrchr
42 /* Search no more than N bytes of S for C. */
44 __memrchr (void const *s, int c_in, size_t n)
46 const unsigned char *char_ptr;
47 const unsigned long int *longword_ptr;
48 unsigned long int longword, magic_bits, charmask;
52 c = (unsigned char) c_in;
54 /* Handle the last few characters by reading one character at a time.
55 Do this until CHAR_PTR is aligned on a longword boundary. */
56 for (char_ptr = (const unsigned char *) s + n;
57 n > 0 && (size_t) char_ptr % sizeof longword != 0;
60 return (void *) char_ptr;
62 /* All these elucidatory comments refer to 4-byte longwords,
63 but the theory applies equally well to any size longwords. */
65 longword_ptr = (const unsigned long int *) char_ptr;
67 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
68 the "holes." Note that there is a hole just to the left of
69 each byte, with an extra at the end:
71 bits: 01111110 11111110 11111110 11111111
72 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
74 The 1-bits make sure that carries propagate to the next 0-bit.
75 The 0-bits provide holes for carries to fall into. */
77 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
78 Set CHARMASK to be a longword, each of whose bytes is C. */
80 magic_bits = 0xfefefefe;
81 charmask = c | (c << 8);
82 charmask |= charmask << 16;
83 #if 0xffffffffU < ULONG_MAX
84 magic_bits |= magic_bits << 32;
85 charmask |= charmask << 32;
86 if (8 < sizeof longword)
87 for (i = 64; i < sizeof longword * 8; i *= 2)
89 magic_bits |= magic_bits << i;
90 charmask |= charmask << i;
93 magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
95 /* Instead of the traditional loop which tests each character,
96 we will test a longword at a time. The tricky part is testing
97 if *any of the four* bytes in the longword in question are zero. */
98 while (n >= sizeof longword)
100 /* We tentatively exit the loop if adding MAGIC_BITS to
101 LONGWORD fails to change any of the hole bits of LONGWORD.
103 1) Is this safe? Will it catch all the zero bytes?
104 Suppose there is a byte with all zeros. Any carry bits
105 propagating from its left will fall into the hole at its
106 least significant bit and stop. Since there will be no
107 carry from its most significant bit, the LSB of the
108 byte to the left will be unchanged, and the zero will be
111 2) Is this worthwhile? Will it ignore everything except
112 zero bytes? Suppose every byte of LONGWORD has a bit set
113 somewhere. There will be a carry into bit 8. If bit 8
114 is set, this will carry into bit 16. If bit 8 is clear,
115 one of bits 9-15 must be set, so there will be a carry
116 into bit 16. Similarly, there will be a carry into bit
117 24. If one of bits 24-30 is set, there will be a carry
118 into bit 31, so all of the hole bits will be changed.
120 The one misfire occurs when bits 24-30 are clear and bit
121 31 is set; in this case, the hole at bit 31 is not
122 changed. If we had access to the processor carry flag,
123 we could close this loophole by putting the fourth hole
126 So it ignores everything except 128's, when they're aligned
129 3) But wait! Aren't we looking for C, not zero?
130 Good point. So what we do is XOR LONGWORD with a longword,
131 each of whose bytes is C. This turns each byte that is C
134 longword = *--longword_ptr ^ charmask;
136 /* Add MAGIC_BITS to LONGWORD. */
137 if ((((longword + magic_bits)
139 /* Set those bits that were unchanged by the addition. */
142 /* Look at only the hole bits. If any of the hole bits
143 are unchanged, most likely one of the bytes was a
147 /* Which of the bytes was C? If none of them were, it was
148 a misfire; continue the search. */
150 const unsigned char *cp = (const unsigned char *) longword_ptr;
152 if (8 < sizeof longword)
153 for (i = sizeof longword - 1; 8 <= i; i--)
155 return (void *) &cp[i];
156 if (7 < sizeof longword && cp[7] == c)
157 return (void *) &cp[7];
158 if (6 < sizeof longword && cp[6] == c)
159 return (void *) &cp[6];
160 if (5 < sizeof longword && cp[5] == c)
161 return (void *) &cp[5];
162 if (4 < sizeof longword && cp[4] == c)
163 return (void *) &cp[4];
165 return (void *) &cp[3];
167 return (void *) &cp[2];
169 return (void *) &cp[1];
174 n -= sizeof longword;
177 char_ptr = (const unsigned char *) longword_ptr;
181 if (*--char_ptr == c)
182 return (void *) char_ptr;
188 weak_alias (__memrchr, memrchr)