1/* Copyright (C) 1991, 1993, 1996, 1997, 1999, 2000, 2003, 2004, 2006, 2008,
2   2009, 2010 Free Software Foundation, Inc.
3
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 in glibc by Roland McGrath (roland@ai.mit.edu).
9   Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
10
11This program is free software: you can redistribute it and/or modify it
12under the terms of the GNU General Public License as published by the
13Free Software Foundation; either version 3 of the License, or any
14later version.
15
16This program is distributed in the hope that it will be useful,
17but WITHOUT ANY WARRANTY; without even the implied warranty of
18MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
19GNU General Public License for more details.
20
21You should have received a copy of the GNU General Public License
22along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
23
24#include <config.h>
25
26#include "memchr2.h"
27
28#include <limits.h>
29#include <stdint.h>
30#include <string.h>
31
32/* Return the first address of either C1 or C2 (treated as unsigned
33   char) that occurs within N bytes of the memory region S.  If
34   neither byte appears, return NULL.  */
35void *
36memchr2 (void const *s, int c1_in, int c2_in, size_t n)
37{
38  /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
39     long instead of a 64-bit uintmax_t tends to give better
40     performance.  On 64-bit hardware, unsigned long is generally 64
41     bits already.  Change this typedef to experiment with
42     performance.  */
43  typedef unsigned long int longword;
44
45  const unsigned char *char_ptr;
46  const longword *longword_ptr;
47  longword repeated_one;
48  longword repeated_c1;
49  longword repeated_c2;
50  unsigned char c1;
51  unsigned char c2;
52
53  c1 = (unsigned char) c1_in;
54  c2 = (unsigned char) c2_in;
55
56  if (c1 == c2)
57    return memchr (s, c1, n);
58
59  /* Handle the first few bytes by reading one byte at a time.
60     Do this until CHAR_PTR is aligned on a longword boundary.  */
61  for (char_ptr = (const unsigned char *) s;
62       n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
63       --n, ++char_ptr)
64    if (*char_ptr == c1 || *char_ptr == c2)
65      return (void *) char_ptr;
66
67  longword_ptr = (const longword *) char_ptr;
68
69  /* All these elucidatory comments refer to 4-byte longwords,
70     but the theory applies equally well to any size longwords.  */
71
72  /* Compute auxiliary longword values:
73     repeated_one is a value which has a 1 in every byte.
74     repeated_c1 has c1 in every byte.
75     repeated_c2 has c2 in every byte.  */
76  repeated_one = 0x01010101;
77  repeated_c1 = c1 | (c1 << 8);
78  repeated_c2 = c2 | (c2 << 8);
79  repeated_c1 |= repeated_c1 << 16;
80  repeated_c2 |= repeated_c2 << 16;
81  if (0xffffffffU < (longword) -1)
82    {
83      repeated_one |= repeated_one << 31 << 1;
84      repeated_c1 |= repeated_c1 << 31 << 1;
85      repeated_c2 |= repeated_c2 << 31 << 1;
86      if (8 < sizeof (longword))
87        {
88          size_t i;
89
90          for (i = 64; i < sizeof (longword) * 8; i *= 2)
91            {
92              repeated_one |= repeated_one << i;
93              repeated_c1 |= repeated_c1 << i;
94              repeated_c2 |= repeated_c2 << i;
95            }
96        }
97    }
98
99  /* Instead of the traditional loop which tests each byte, we will test a
100     longword at a time.  The tricky part is testing if *any of the four*
101     bytes in the longword in question are equal to c1 or c2.  We first use
102     an xor with repeated_c1 and repeated_c2, respectively.  This reduces
103     the task to testing whether *any of the four* bytes in longword1 or
104     longword2 is zero.
105
106     Let's consider longword1.  We compute tmp1 =
107       ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
108     That is, we perform the following operations:
109       1. Subtract repeated_one.
110       2. & ~longword1.
111       3. & a mask consisting of 0x80 in every byte.
112     Consider what happens in each byte:
113       - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
114         and step 3 transforms it into 0x80.  A carry can also be propagated
115         to more significant bytes.
116       - If a byte of longword1 is nonzero, let its lowest 1 bit be at
117         position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
118         the byte ends in a single bit of value 0 and k bits of value 1.
119         After step 2, the result is just k bits of value 1: 2^k - 1.  After
120         step 3, the result is 0.  And no carry is produced.
121     So, if longword1 has only non-zero bytes, tmp1 is zero.
122     Whereas if longword1 has a zero byte, call j the position of the least
123     significant zero byte.  Then the result has a zero at positions 0, ...,
124     j-1 and a 0x80 at position j.  We cannot predict the result at the more
125     significant bytes (positions j+1..3), but it does not matter since we
126     already have a non-zero bit at position 8*j+7.
127
128     Similary, we compute tmp2 =
129       ((longword2 - repeated_one) & ~longword2) & (repeated_one << 7).
130
131     The test whether any byte in longword1 or longword2 is zero is equivalent
132     to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine
133     this into a single test, whether (tmp1 | tmp2) is nonzero.  */
134
135  while (n >= sizeof (longword))
136    {
137      longword longword1 = *longword_ptr ^ repeated_c1;
138      longword longword2 = *longword_ptr ^ repeated_c2;
139
140      if (((((longword1 - repeated_one) & ~longword1)
141            | ((longword2 - repeated_one) & ~longword2))
142           & (repeated_one << 7)) != 0)
143        break;
144      longword_ptr++;
145      n -= sizeof (longword);
146    }
147
148  char_ptr = (const unsigned char *) longword_ptr;
149
150  /* At this point, we know that either n < sizeof (longword), or one of the
151     sizeof (longword) bytes starting at char_ptr is == c1 or == c2.  On
152     little-endian machines, we could determine the first such byte without
153     any further memory accesses, just by looking at the (tmp1 | tmp2) result
154     from the last loop iteration.  But this does not work on big-endian
155     machines.  Choose code that works in both cases.  */
156
157  for (; n > 0; --n, ++char_ptr)
158    {
159      if (*char_ptr == c1 || *char_ptr == c2)
160        return (void *) char_ptr;
161    }
162
163  return NULL;
164}
165