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 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 2, or (at your option)
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 along
23 with this program; if not, write to the Free Software Foundation,
24 Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA. */
32 # define reg_char char
41 # define __memrchr memrchr
44 /* Search no more than N bytes of S for C. */
46 __memrchr (void const *s, int c_in, size_t n)
48 const unsigned char *char_ptr;
49 const unsigned long int *longword_ptr;
50 unsigned long int longword, magic_bits, charmask;
54 c = (unsigned char) c_in;
56 /* Handle the last few characters by reading one character at a time.
57 Do this until CHAR_PTR is aligned on a longword boundary. */
58 for (char_ptr = (const unsigned char *) s + n;
59 n > 0 && (size_t) char_ptr % sizeof longword != 0;
62 return (void *) char_ptr;
64 /* All these elucidatory comments refer to 4-byte longwords,
65 but the theory applies equally well to any size longwords. */
67 longword_ptr = (const unsigned long int *) char_ptr;
69 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
70 the "holes." Note that there is a hole just to the left of
71 each byte, with an extra at the end:
73 bits: 01111110 11111110 11111110 11111111
74 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
76 The 1-bits make sure that carries propagate to the next 0-bit.
77 The 0-bits provide holes for carries to fall into. */
79 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
80 Set CHARMASK to be a longword, each of whose bytes is C. */
82 magic_bits = 0xfefefefe;
83 charmask = c | (c << 8);
84 charmask |= charmask << 16;
85 #if 0xffffffffU < ULONG_MAX
86 magic_bits |= magic_bits << 32;
87 charmask |= charmask << 32;
88 if (8 < sizeof longword)
89 for (i = 64; i < sizeof longword * 8; i *= 2)
91 magic_bits |= magic_bits << i;
92 charmask |= charmask << i;
95 magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
97 /* Instead of the traditional loop which tests each character,
98 we will test a longword at a time. The tricky part is testing
99 if *any of the four* bytes in the longword in question are zero. */
100 while (n >= sizeof longword)
102 /* We tentatively exit the loop if adding MAGIC_BITS to
103 LONGWORD fails to change any of the hole bits of LONGWORD.
105 1) Is this safe? Will it catch all the zero bytes?
106 Suppose there is a byte with all zeros. Any carry bits
107 propagating from its left will fall into the hole at its
108 least significant bit and stop. Since there will be no
109 carry from its most significant bit, the LSB of the
110 byte to the left will be unchanged, and the zero will be
113 2) Is this worthwhile? Will it ignore everything except
114 zero bytes? Suppose every byte of LONGWORD has a bit set
115 somewhere. There will be a carry into bit 8. If bit 8
116 is set, this will carry into bit 16. If bit 8 is clear,
117 one of bits 9-15 must be set, so there will be a carry
118 into bit 16. Similarly, there will be a carry into bit
119 24. If one of bits 24-30 is set, there will be a carry
120 into bit 31, so all of the hole bits will be changed.
122 The one misfire occurs when bits 24-30 are clear and bit
123 31 is set; in this case, the hole at bit 31 is not
124 changed. If we had access to the processor carry flag,
125 we could close this loophole by putting the fourth hole
128 So it ignores everything except 128's, when they're aligned
131 3) But wait! Aren't we looking for C, not zero?
132 Good point. So what we do is XOR LONGWORD with a longword,
133 each of whose bytes is C. This turns each byte that is C
136 longword = *--longword_ptr ^ charmask;
138 /* Add MAGIC_BITS to LONGWORD. */
139 if ((((longword + magic_bits)
141 /* Set those bits that were unchanged by the addition. */
144 /* Look at only the hole bits. If any of the hole bits
145 are unchanged, most likely one of the bytes was a
149 /* Which of the bytes was C? If none of them were, it was
150 a misfire; continue the search. */
152 const unsigned char *cp = (const unsigned char *) longword_ptr;
154 if (8 < sizeof longword)
155 for (i = sizeof longword - 1; 8 <= i; i--)
157 return (void *) &cp[i];
158 if (7 < sizeof longword && cp[7] == c)
159 return (void *) &cp[7];
160 if (6 < sizeof longword && cp[6] == c)
161 return (void *) &cp[6];
162 if (5 < sizeof longword && cp[5] == c)
163 return (void *) &cp[5];
164 if (4 < sizeof longword && cp[4] == c)
165 return (void *) &cp[4];
167 return (void *) &cp[3];
169 return (void *) &cp[2];
171 return (void *) &cp[1];
176 n -= sizeof longword;
179 char_ptr = (const unsigned char *) longword_ptr;
183 if (*--char_ptr == c)
184 return (void *) char_ptr;
190 weak_alias (__memrchr, memrchr)