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 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. */
30 # define reg_char char
40 # define __memrchr memrchr
43 /* Search no more than N bytes of S for C. */
45 __memrchr (void const *s, int c_in, size_t n)
47 const unsigned char *char_ptr;
48 const unsigned long int *longword_ptr;
49 unsigned long int longword, magic_bits, charmask;
53 c = (unsigned char) c_in;
55 /* Handle the last few characters by reading one character at a time.
56 Do this until CHAR_PTR is aligned on a longword boundary. */
57 for (char_ptr = (const unsigned char *) s + n;
58 n > 0 && (size_t) char_ptr % sizeof longword != 0;
61 return (void *) char_ptr;
63 /* All these elucidatory comments refer to 4-byte longwords,
64 but the theory applies equally well to any size longwords. */
66 longword_ptr = (const unsigned long int *) char_ptr;
68 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
69 the "holes." Note that there is a hole just to the left of
70 each byte, with an extra at the end:
72 bits: 01111110 11111110 11111110 11111111
73 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
75 The 1-bits make sure that carries propagate to the next 0-bit.
76 The 0-bits provide holes for carries to fall into. */
78 /* Set MAGIC_BITS to be this pattern of 1 and 0 bits.
79 Set CHARMASK to be a longword, each of whose bytes is C. */
81 magic_bits = 0xfefefefe;
82 charmask = c | (c << 8);
83 charmask |= charmask << 16;
84 #if 0xffffffffU < ULONG_MAX
85 magic_bits |= magic_bits << 32;
86 charmask |= charmask << 32;
87 if (8 < sizeof longword)
88 for (i = 64; i < sizeof longword * 8; i *= 2)
90 magic_bits |= magic_bits << i;
91 charmask |= charmask << i;
94 magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
96 /* Instead of the traditional loop which tests each character,
97 we will test a longword at a time. The tricky part is testing
98 if *any of the four* bytes in the longword in question are zero. */
99 while (n >= sizeof longword)
101 /* We tentatively exit the loop if adding MAGIC_BITS to
102 LONGWORD fails to change any of the hole bits of LONGWORD.
104 1) Is this safe? Will it catch all the zero bytes?
105 Suppose there is a byte with all zeros. Any carry bits
106 propagating from its left will fall into the hole at its
107 least significant bit and stop. Since there will be no
108 carry from its most significant bit, the LSB of the
109 byte to the left will be unchanged, and the zero will be
112 2) Is this worthwhile? Will it ignore everything except
113 zero bytes? Suppose every byte of LONGWORD has a bit set
114 somewhere. There will be a carry into bit 8. If bit 8
115 is set, this will carry into bit 16. If bit 8 is clear,
116 one of bits 9-15 must be set, so there will be a carry
117 into bit 16. Similarly, there will be a carry into bit
118 24. If one of bits 24-30 is set, there will be a carry
119 into bit 31, so all of the hole bits will be changed.
121 The one misfire occurs when bits 24-30 are clear and bit
122 31 is set; in this case, the hole at bit 31 is not
123 changed. If we had access to the processor carry flag,
124 we could close this loophole by putting the fourth hole
127 So it ignores everything except 128's, when they're aligned
130 3) But wait! Aren't we looking for C, not zero?
131 Good point. So what we do is XOR LONGWORD with a longword,
132 each of whose bytes is C. This turns each byte that is C
135 longword = *--longword_ptr ^ charmask;
137 /* Add MAGIC_BITS to LONGWORD. */
138 if ((((longword + magic_bits)
140 /* Set those bits that were unchanged by the addition. */
143 /* Look at only the hole bits. If any of the hole bits
144 are unchanged, most likely one of the bytes was a
148 /* Which of the bytes was C? If none of them were, it was
149 a misfire; continue the search. */
151 const unsigned char *cp = (const unsigned char *) longword_ptr;
153 if (8 < sizeof longword)
154 for (i = sizeof longword - 1; 8 <= i; i--)
156 return (void *) &cp[i];
157 if (7 < sizeof longword && cp[7] == c)
158 return (void *) &cp[7];
159 if (6 < sizeof longword && cp[6] == c)
160 return (void *) &cp[6];
161 if (5 < sizeof longword && cp[5] == c)
162 return (void *) &cp[5];
163 if (4 < sizeof longword && cp[4] == c)
164 return (void *) &cp[4];
166 return (void *) &cp[3];
168 return (void *) &cp[2];
170 return (void *) &cp[1];
175 n -= sizeof longword;
178 char_ptr = (const unsigned char *) longword_ptr;
182 if (*--char_ptr == c)
183 return (void *) char_ptr;
189 weak_alias (__memrchr, memrchr)