1 /* memrchr -- find the last occurrence of a byte in a memory block
2 Copyright (C) 1991, 1993, 1996, 1997, 1999 Free Software Foundation, Inc.
3 Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
4 with help from Dan Sahlin (dan@sics.se) and
5 commentary by Jim Blandy (jimb@ai.mit.edu);
6 adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
7 and implemented by Roland McGrath (roland@ai.mit.edu).
9 The GNU C Library is free software; you can redistribute it and/or
10 modify it under the terms of the GNU Library General Public License as
11 published by the Free Software Foundation; either version 2 of the
12 License, or (at your option) any later version.
14 The GNU C Library is distributed in the hope that it will be useful,
15 but WITHOUT ANY WARRANTY; without even the implied warranty of
16 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
17 Library General Public License for more details.
19 You should have received a copy of the GNU Library General Public
20 License along with the GNU C Library; see the file COPYING.LIB. If not,
21 write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
22 Boston, MA 02111-1307, USA. */
29 #if defined (__cplusplus) || (defined (__STDC__) && __STDC__)
30 # define __ptr_t void *
31 #else /* Not C++ or ANSI C. */
32 # define __ptr_t char *
33 #endif /* C++ or ANSI C. */
39 # define reg_char char
42 #if defined (HAVE_LIMITS_H) || defined (_LIBC)
46 #define LONG_MAX_32_BITS 2147483647
49 # define LONG_MAX LONG_MAX_32_BITS
52 #include <sys/types.h>
57 /* Search no more than N bytes of S for C. */
59 __memrchr (s, c_in, n)
64 const unsigned char *char_ptr;
65 const unsigned long int *longword_ptr;
66 unsigned long int longword, magic_bits, charmask;
69 c = (unsigned char) c_in;
71 /* Handle the last few characters by reading one character at a time.
72 Do this until CHAR_PTR is aligned on a longword boundary. */
73 for (char_ptr = (const unsigned char *) s + n;
74 n > 0 && ((unsigned long int) char_ptr
75 & (sizeof (longword) - 1)) != 0;
78 return (__ptr_t) char_ptr;
80 /* All these elucidatory comments refer to 4-byte longwords,
81 but the theory applies equally well to 8-byte longwords. */
83 longword_ptr = (unsigned long int *) char_ptr;
85 /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
86 the "holes." Note that there is a hole just to the left of
87 each byte, with an extra at the end:
89 bits: 01111110 11111110 11111110 11111111
90 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
92 The 1-bits make sure that carries propagate to the next 0-bit.
93 The 0-bits provide holes for carries to fall into. */
95 if (sizeof (longword) != 4 && sizeof (longword) != 8)
98 #if LONG_MAX <= LONG_MAX_32_BITS
99 magic_bits = 0x7efefeff;
101 magic_bits = ((unsigned long int) 0x7efefefe << 32) | 0xfefefeff;
104 /* Set up a longword, each of whose bytes is C. */
105 charmask = c | (c << 8);
106 charmask |= charmask << 16;
107 #if LONG_MAX > LONG_MAX_32_BITS
108 charmask |= charmask << 32;
111 /* Instead of the traditional loop which tests each character,
112 we will test a longword at a time. The tricky part is testing
113 if *any of the four* bytes in the longword in question are zero. */
114 while (n >= sizeof (longword))
116 /* We tentatively exit the loop if adding MAGIC_BITS to
117 LONGWORD fails to change any of the hole bits of LONGWORD.
119 1) Is this safe? Will it catch all the zero bytes?
120 Suppose there is a byte with all zeros. Any carry bits
121 propagating from its left will fall into the hole at its
122 least significant bit and stop. Since there will be no
123 carry from its most significant bit, the LSB of the
124 byte to the left will be unchanged, and the zero will be
127 2) Is this worthwhile? Will it ignore everything except
128 zero bytes? Suppose every byte of LONGWORD has a bit set
129 somewhere. There will be a carry into bit 8. If bit 8
130 is set, this will carry into bit 16. If bit 8 is clear,
131 one of bits 9-15 must be set, so there will be a carry
132 into bit 16. Similarly, there will be a carry into bit
133 24. If one of bits 24-30 is set, there will be a carry
134 into bit 31, so all of the hole bits will be changed.
136 The one misfire occurs when bits 24-30 are clear and bit
137 31 is set; in this case, the hole at bit 31 is not
138 changed. If we had access to the processor carry flag,
139 we could close this loophole by putting the fourth hole
142 So it ignores everything except 128's, when they're aligned
145 3) But wait! Aren't we looking for C, not zero?
146 Good point. So what we do is XOR LONGWORD with a longword,
147 each of whose bytes is C. This turns each byte that is C
150 longword = *--longword_ptr ^ charmask;
152 /* Add MAGIC_BITS to LONGWORD. */
153 if ((((longword + magic_bits)
155 /* Set those bits that were unchanged by the addition. */
158 /* Look at only the hole bits. If any of the hole bits
159 are unchanged, most likely one of the bytes was a
163 /* Which of the bytes was C? If none of them were, it was
164 a misfire; continue the search. */
166 const unsigned char *cp = (const unsigned char *) longword_ptr;
168 #if LONG_MAX > 2147483647
170 return (__ptr_t) &cp[7];
172 return (__ptr_t) &cp[6];
174 return (__ptr_t) &cp[5];
176 return (__ptr_t) &cp[4];
179 return (__ptr_t) &cp[3];
181 return (__ptr_t) &cp[2];
183 return (__ptr_t) &cp[1];
188 n -= sizeof (longword);
191 char_ptr = (const unsigned char *) longword_ptr;
195 if (*--char_ptr == c)
196 return (__ptr_t) char_ptr;
201 weak_alias (__memrchr, memrchr)