1 /* memrchr -- find the last occurrence of a byte in a memory block
3 Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003 Free
4 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., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA. */
36 # define reg_char char
41 #define LONG_MAX_32_BITS 2147483647
43 #include <sys/types.h>
49 # define __memrchr memrchr
52 /* Search no more than N bytes of S for C. */
54 __memrchr (void const *s
, int c_in
, size_t n
)
56 const unsigned char *char_ptr
;
57 const unsigned long int *longword_ptr
;
58 unsigned long int longword
, magic_bits
, charmask
;
61 c
= (unsigned char) c_in
;
63 /* Handle the last few characters by reading one character at a time.
64 Do this until CHAR_PTR is aligned on a longword boundary. */
65 for (char_ptr
= (const unsigned char *) s
+ n
;
66 n
> 0 && ((unsigned long int) char_ptr
67 & (sizeof (longword
) - 1)) != 0;
70 return (void *) char_ptr
;
72 /* All these elucidatory comments refer to 4-byte longwords,
73 but the theory applies equally well to 8-byte longwords. */
75 longword_ptr
= (unsigned long int *) char_ptr
;
77 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
78 the "holes." Note that there is a hole just to the left of
79 each byte, with an extra at the end:
81 bits: 01111110 11111110 11111110 11111111
82 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
84 The 1-bits make sure that carries propagate to the next 0-bit.
85 The 0-bits provide holes for carries to fall into. */
87 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
90 #if LONG_MAX <= LONG_MAX_32_BITS
91 magic_bits
= 0x7efefeff;
93 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
96 /* Set up a longword, each of whose bytes is C. */
97 charmask
= c
| (c
<< 8);
98 charmask
|= charmask
<< 16;
99 #if LONG_MAX > LONG_MAX_32_BITS
100 charmask
|= charmask
<< 32;
103 /* Instead of the traditional loop which tests each character,
104 we will test a longword at a time. The tricky part is testing
105 if *any of the four* bytes in the longword in question are zero. */
106 while (n
>= sizeof (longword
))
108 /* We tentatively exit the loop if adding MAGIC_BITS to
109 LONGWORD fails to change any of the hole bits of LONGWORD.
111 1) Is this safe? Will it catch all the zero bytes?
112 Suppose there is a byte with all zeros. Any carry bits
113 propagating from its left will fall into the hole at its
114 least significant bit and stop. Since there will be no
115 carry from its most significant bit, the LSB of the
116 byte to the left will be unchanged, and the zero will be
119 2) Is this worthwhile? Will it ignore everything except
120 zero bytes? Suppose every byte of LONGWORD has a bit set
121 somewhere. There will be a carry into bit 8. If bit 8
122 is set, this will carry into bit 16. If bit 8 is clear,
123 one of bits 9-15 must be set, so there will be a carry
124 into bit 16. Similarly, there will be a carry into bit
125 24. If one of bits 24-30 is set, there will be a carry
126 into bit 31, so all of the hole bits will be changed.
128 The one misfire occurs when bits 24-30 are clear and bit
129 31 is set; in this case, the hole at bit 31 is not
130 changed. If we had access to the processor carry flag,
131 we could close this loophole by putting the fourth hole
134 So it ignores everything except 128's, when they're aligned
137 3) But wait! Aren't we looking for C, not zero?
138 Good point. So what we do is XOR LONGWORD with a longword,
139 each of whose bytes is C. This turns each byte that is C
142 longword
= *--longword_ptr
^ charmask
;
144 /* Add MAGIC_BITS to LONGWORD. */
145 if ((((longword
+ magic_bits
)
147 /* Set those bits that were unchanged by the addition. */
150 /* Look at only the hole bits. If any of the hole bits
151 are unchanged, most likely one of the bytes was a
155 /* Which of the bytes was C? If none of them were, it was
156 a misfire; continue the search. */
158 const unsigned char *cp
= (const unsigned char *) longword_ptr
;
160 #if LONG_MAX > 2147483647
162 return (void *) &cp
[7];
164 return (void *) &cp
[6];
166 return (void *) &cp
[5];
168 return (void *) &cp
[4];
171 return (void *) &cp
[3];
173 return (void *) &cp
[2];
175 return (void *) &cp
[1];
180 n
-= sizeof (longword
);
183 char_ptr
= (const unsigned char *) longword_ptr
;
187 if (*--char_ptr
== c
)
188 return (void *) char_ptr
;
194 weak_alias (__memrchr
, memrchr
)