1/* memrchr -- find the last occurrence of a byte in a memory block
2
3   Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2005,
4   2006, 2007 Free Software Foundation, Inc.
5
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).
11
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)
15   any later version.
16
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.
21
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.  */
25
26#if defined _LIBC
27# include <memcopy.h>
28#else
29# include <config.h>
30# define reg_char char
31#endif
32
33#include <string.h>
34#include <limits.h>
35
36#undef __memrchr
37#undef memrchr
38
39#ifndef weak_alias
40# define __memrchr memrchr
41#endif
42
43/* Search no more than N bytes of S for C.  */
44void *
45__memrchr (void const *s, int c_in, size_t n)
46{
47  const unsigned char *char_ptr;
48  const unsigned long int *longword_ptr;
49  unsigned long int longword, magic_bits, charmask;
50  unsigned reg_char c;
51  int i;
52
53  c = (unsigned char) c_in;
54
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;
59       --n)
60    if (*--char_ptr == c)
61      return (void *) char_ptr;
62
63  /* All these elucidatory comments refer to 4-byte longwords,
64     but the theory applies equally well to any size longwords.  */
65
66  longword_ptr = (const unsigned long int *) char_ptr;
67
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:
71
72     bits:  01111110 11111110 11111110 11111111
73     bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
74
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.  */
77
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.  */
80
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)
89      {
90	magic_bits |= magic_bits << i;
91	charmask |= charmask << i;
92      }
93#endif
94  magic_bits = (ULONG_MAX >> 1) & (magic_bits | 1);
95
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)
100    {
101      /* We tentatively exit the loop if adding MAGIC_BITS to
102	 LONGWORD fails to change any of the hole bits of LONGWORD.
103
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
110	 detected.
111
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.
120
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
125	 at bit 32!
126
127	 So it ignores everything except 128's, when they're aligned
128	 properly.
129
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
133	 into a zero.  */
134
135      longword = *--longword_ptr ^ charmask;
136
137      /* Add MAGIC_BITS to LONGWORD.  */
138      if ((((longword + magic_bits)
139
140	    /* Set those bits that were unchanged by the addition.  */
141	    ^ ~longword)
142
143	   /* Look at only the hole bits.  If any of the hole bits
144	      are unchanged, most likely one of the bytes was a
145	      zero.  */
146	   & ~magic_bits) != 0)
147	{
148	  /* Which of the bytes was C?  If none of them were, it was
149	     a misfire; continue the search.  */
150
151	  const unsigned char *cp = (const unsigned char *) longword_ptr;
152
153	  if (8 < sizeof longword)
154	    for (i = sizeof longword - 1; 8 <= i; i--)
155	      if (cp[i] == c)
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];
165	  if (cp[3] == c)
166	    return (void *) &cp[3];
167	  if (cp[2] == c)
168	    return (void *) &cp[2];
169	  if (cp[1] == c)
170	    return (void *) &cp[1];
171	  if (cp[0] == c)
172	    return (void *) cp;
173	}
174
175      n -= sizeof longword;
176    }
177
178  char_ptr = (const unsigned char *) longword_ptr;
179
180  while (n-- > 0)
181    {
182      if (*--char_ptr == c)
183	return (void *) char_ptr;
184    }
185
186  return 0;
187}
188#ifdef weak_alias
189weak_alias (__memrchr, memrchr)
190#endif
191