1 /* memrchr -- find the last occurrence of a byte in a memory block
2 Copyright (C) 1991-2015 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, see
22 <http://www.gnu.org/licenses/>. */
31 #define __ptr_t void *
38 #if defined HAVE_LIMITS_H || defined _LIBC
42 #define LONG_MAX_32_BITS 2147483647
45 # define LONG_MAX LONG_MAX_32_BITS
48 #include <sys/types.h>
54 # define __memrchr memrchr
57 /* Search no more than N bytes of S for C. */
69 const unsigned char *char_ptr
;
70 const unsigned long int *longword_ptr
;
71 unsigned long int longword
, magic_bits
, charmask
;
74 c
= (unsigned char) c_in
;
76 /* Handle the last few characters by reading one character at a time.
77 Do this until CHAR_PTR is aligned on a longword boundary. */
78 for (char_ptr
= (const unsigned char *) s
+ n
;
79 n
> 0 && ((unsigned long int) char_ptr
80 & (sizeof (longword
) - 1)) != 0;
83 return (__ptr_t
) char_ptr
;
85 /* All these elucidatory comments refer to 4-byte longwords,
86 but the theory applies equally well to 8-byte longwords. */
88 longword_ptr
= (const unsigned long int *) char_ptr
;
90 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
91 the "holes." Note that there is a hole just to the left of
92 each byte, with an extra at the end:
94 bits: 01111110 11111110 11111110 11111111
95 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
97 The 1-bits make sure that carries propagate to the next 0-bit.
98 The 0-bits provide holes for carries to fall into. */
100 if (sizeof (longword
) != 4 && sizeof (longword
) != 8)
103 #if LONG_MAX <= LONG_MAX_32_BITS
104 magic_bits
= 0x7efefeff;
106 magic_bits
= ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
109 /* Set up a longword, each of whose bytes is C. */
110 charmask
= c
| (c
<< 8);
111 charmask
|= charmask
<< 16;
112 #if LONG_MAX > LONG_MAX_32_BITS
113 charmask
|= charmask
<< 32;
116 /* Instead of the traditional loop which tests each character,
117 we will test a longword at a time. The tricky part is testing
118 if *any of the four* bytes in the longword in question are zero. */
119 while (n
>= sizeof (longword
))
121 /* We tentatively exit the loop if adding MAGIC_BITS to
122 LONGWORD fails to change any of the hole bits of LONGWORD.
124 1) Is this safe? Will it catch all the zero bytes?
125 Suppose there is a byte with all zeros. Any carry bits
126 propagating from its left will fall into the hole at its
127 least significant bit and stop. Since there will be no
128 carry from its most significant bit, the LSB of the
129 byte to the left will be unchanged, and the zero will be
132 2) Is this worthwhile? Will it ignore everything except
133 zero bytes? Suppose every byte of LONGWORD has a bit set
134 somewhere. There will be a carry into bit 8. If bit 8
135 is set, this will carry into bit 16. If bit 8 is clear,
136 one of bits 9-15 must be set, so there will be a carry
137 into bit 16. Similarly, there will be a carry into bit
138 24. If one of bits 24-30 is set, there will be a carry
139 into bit 31, so all of the hole bits will be changed.
141 The one misfire occurs when bits 24-30 are clear and bit
142 31 is set; in this case, the hole at bit 31 is not
143 changed. If we had access to the processor carry flag,
144 we could close this loophole by putting the fourth hole
147 So it ignores everything except 128's, when they're aligned
150 3) But wait! Aren't we looking for C, not zero?
151 Good point. So what we do is XOR LONGWORD with a longword,
152 each of whose bytes is C. This turns each byte that is C
155 longword
= *--longword_ptr
^ charmask
;
157 /* Add MAGIC_BITS to LONGWORD. */
158 if ((((longword
+ magic_bits
)
160 /* Set those bits that were unchanged by the addition. */
163 /* Look at only the hole bits. If any of the hole bits
164 are unchanged, most likely one of the bytes was a
168 /* Which of the bytes was C? If none of them were, it was
169 a misfire; continue the search. */
171 const unsigned char *cp
= (const unsigned char *) longword_ptr
;
173 #if LONG_MAX > 2147483647
175 return (__ptr_t
) &cp
[7];
177 return (__ptr_t
) &cp
[6];
179 return (__ptr_t
) &cp
[5];
181 return (__ptr_t
) &cp
[4];
184 return (__ptr_t
) &cp
[3];
186 return (__ptr_t
) &cp
[2];
188 return (__ptr_t
) &cp
[1];
193 n
-= sizeof (longword
);
196 char_ptr
= (const unsigned char *) longword_ptr
;
200 if (*--char_ptr
== c
)
201 return (__ptr_t
) char_ptr
;
208 weak_alias (__memrchr
, memrchr
)