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, write to the Free
22 Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA
32 #if defined __cplusplus || (defined __STDC__ && __STDC__)
33 # define __ptr_t void *
34 #else /* Not C++ or ANSI C. */
35 # define __ptr_t char *
36 #endif /* C++ or ANSI C. */
42 # define reg_char char
45 #if defined HAVE_LIMITS_H || defined _LIBC
49 #define LONG_MAX_32_BITS 2147483647
52 # define LONG_MAX LONG_MAX_32_BITS
55 #include <sys/types.h>
61 # define __memrchr memrchr
64 /* Search no more than N bytes of S for C. */
66 __memrchr (s
, c_in
, n
)
71 const unsigned char *char_ptr
;
72 const unsigned long int *longword_ptr
;
73 unsigned long int longword
, magic_bits
, charmask
;
76 c
= (unsigned char) c_in
;
78 /* Handle the last few characters by reading one character at a time.
79 Do this until CHAR_PTR is aligned on a longword boundary. */
80 for (char_ptr
= (const unsigned char *) s
+ n
;
81 n
> 0 && ((unsigned long int) char_ptr
82 & (sizeof (longword
) - 1)) != 0;
85 return (__ptr_t
) char_ptr
;
87 /* All these elucidatory comments refer to 4-byte longwords,
88 but the theory applies equally well to 8-byte longwords. */
90 longword_ptr
= (const unsigned long int *) char_ptr
;
92 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
93 the "holes." Note that there is a hole just to the left of
94 each byte, with an extra at the end:
96 bits: 01111110 11111110 11111110 11111111
97 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
99 The 1-bits make sure that carries propagate to the next 0-bit.
100 The 0-bits provide holes for carries to fall into. */
102 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
105 #if LONG_MAX <= LONG_MAX_32_BITS
106 magic_bits
= 0x7efefeff;
108 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
111 /* Set up a longword, each of whose bytes is C. */
112 charmask
= c
| (c
<< 8);
113 charmask
|= charmask
<< 16;
114 #if LONG_MAX > LONG_MAX_32_BITS
115 charmask
|= charmask
<< 32;
118 /* Instead of the traditional loop which tests each character,
119 we will test a longword at a time. The tricky part is testing
120 if *any of the four* bytes in the longword in question are zero. */
121 while (n
>= sizeof (longword
))
123 /* We tentatively exit the loop if adding MAGIC_BITS to
124 LONGWORD fails to change any of the hole bits of LONGWORD.
126 1) Is this safe? Will it catch all the zero bytes?
127 Suppose there is a byte with all zeros. Any carry bits
128 propagating from its left will fall into the hole at its
129 least significant bit and stop. Since there will be no
130 carry from its most significant bit, the LSB of the
131 byte to the left will be unchanged, and the zero will be
134 2) Is this worthwhile? Will it ignore everything except
135 zero bytes? Suppose every byte of LONGWORD has a bit set
136 somewhere. There will be a carry into bit 8. If bit 8
137 is set, this will carry into bit 16. If bit 8 is clear,
138 one of bits 9-15 must be set, so there will be a carry
139 into bit 16. Similarly, there will be a carry into bit
140 24. If one of bits 24-30 is set, there will be a carry
141 into bit 31, so all of the hole bits will be changed.
143 The one misfire occurs when bits 24-30 are clear and bit
144 31 is set; in this case, the hole at bit 31 is not
145 changed. If we had access to the processor carry flag,
146 we could close this loophole by putting the fourth hole
149 So it ignores everything except 128's, when they're aligned
152 3) But wait! Aren't we looking for C, not zero?
153 Good point. So what we do is XOR LONGWORD with a longword,
154 each of whose bytes is C. This turns each byte that is C
157 longword
= *--longword_ptr
^ charmask
;
159 /* Add MAGIC_BITS to LONGWORD. */
160 if ((((longword
+ magic_bits
)
162 /* Set those bits that were unchanged by the addition. */
165 /* Look at only the hole bits. If any of the hole bits
166 are unchanged, most likely one of the bytes was a
170 /* Which of the bytes was C? If none of them were, it was
171 a misfire; continue the search. */
173 const unsigned char *cp
= (const unsigned char *) longword_ptr
;
175 #if LONG_MAX > 2147483647
177 return (__ptr_t
) &cp
[7];
179 return (__ptr_t
) &cp
[6];
181 return (__ptr_t
) &cp
[5];
183 return (__ptr_t
) &cp
[4];
186 return (__ptr_t
) &cp
[3];
188 return (__ptr_t
) &cp
[2];
190 return (__ptr_t
) &cp
[1];
195 n
-= sizeof (longword
);
198 char_ptr
= (const unsigned char *) longword_ptr
;
202 if (*--char_ptr
== c
)
203 return (__ptr_t
) char_ptr
;
209 weak_alias (__memrchr
, memrchr
)