1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2021
2    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 by Roland McGrath (roland@ai.mit.edu).
9 
10 NOTE: The canonical source of this file is maintained with the GNU C Library.
11 Bugs can be reported to bug-glibc@prep.ai.mit.edu.
12 
13 This program is free software: you can redistribute it and/or modify it
14 under the terms of the GNU General Public License as published by the
15 Free Software Foundation; either version 3 of the License, or any
16 later version.
17 
18 This program is distributed in the hope that it will be useful,
19 but WITHOUT ANY WARRANTY; without even the implied warranty of
20 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
21 GNU General Public License for more details.
22 
23 You should have received a copy of the GNU General Public License
24 along with this program.  If not, see <https://www.gnu.org/licenses/>.  */
25 
26 #ifndef _LIBC
27 # include <config.h>
28 #endif
29 
30 #include <string.h>
31 
32 #include <stddef.h>
33 
34 #if defined _LIBC
35 # include <memcopy.h>
36 #else
37 # define reg_char char
38 #endif
39 
40 #include <limits.h>
41 
42 #if HAVE_BP_SYM_H || defined _LIBC
43 # include <bp-sym.h>
44 #else
45 # define BP_SYM(sym) sym
46 #endif
47 
48 #undef __memchr
49 #ifdef _LIBC
50 # undef memchr
51 #endif
52 
53 #ifndef weak_alias
54 # define __memchr memchr
55 #endif
56 
57 /* Search no more than N bytes of S for C.  */
58 void *
__memchr(void const * s,int c_in,size_t n)59 __memchr (void const *s, int c_in, size_t n)
60 {
61   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
62      long instead of a 64-bit uintmax_t tends to give better
63      performance.  On 64-bit hardware, unsigned long is generally 64
64      bits already.  Change this typedef to experiment with
65      performance.  */
66   typedef unsigned long int longword;
67 
68   const unsigned char *char_ptr;
69   const longword *longword_ptr;
70   longword repeated_one;
71   longword repeated_c;
72   unsigned reg_char c;
73 
74   c = (unsigned char) c_in;
75 
76   /* Handle the first few bytes by reading one byte at a time.
77      Do this until CHAR_PTR is aligned on a longword boundary.  */
78   for (char_ptr = (const unsigned char *) s;
79        n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
80        --n, ++char_ptr)
81     if (*char_ptr == c)
82       return (void *) char_ptr;
83 
84   longword_ptr = (const longword *) char_ptr;
85 
86   /* All these elucidatory comments refer to 4-byte longwords,
87      but the theory applies equally well to any size longwords.  */
88 
89   /* Compute auxiliary longword values:
90      repeated_one is a value which has a 1 in every byte.
91      repeated_c has c in every byte.  */
92   repeated_one = 0x01010101;
93   repeated_c = c | (c << 8);
94   repeated_c |= repeated_c << 16;
95   if (0xffffffffU < (longword) -1)
96     {
97       repeated_one |= repeated_one << 31 << 1;
98       repeated_c |= repeated_c << 31 << 1;
99       if (8 < sizeof (longword))
100         {
101           size_t i;
102 
103           for (i = 64; i < sizeof (longword) * 8; i *= 2)
104             {
105               repeated_one |= repeated_one << i;
106               repeated_c |= repeated_c << i;
107             }
108         }
109     }
110 
111   /* Instead of the traditional loop which tests each byte, we will test a
112      longword at a time.  The tricky part is testing if *any of the four*
113      bytes in the longword in question are equal to c.  We first use an xor
114      with repeated_c.  This reduces the task to testing whether *any of the
115      four* bytes in longword1 is zero.
116 
117      We compute tmp =
118        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
119      That is, we perform the following operations:
120        1. Subtract repeated_one.
121        2. & ~longword1.
122        3. & a mask consisting of 0x80 in every byte.
123      Consider what happens in each byte:
124        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
125          and step 3 transforms it into 0x80.  A carry can also be propagated
126          to more significant bytes.
127        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
128          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
129          the byte ends in a single bit of value 0 and k bits of value 1.
130          After step 2, the result is just k bits of value 1: 2^k - 1.  After
131          step 3, the result is 0.  And no carry is produced.
132      So, if longword1 has only non-zero bytes, tmp is zero.
133      Whereas if longword1 has a zero byte, call j the position of the least
134      significant zero byte.  Then the result has a zero at positions 0, ...,
135      j-1 and a 0x80 at position j.  We cannot predict the result at the more
136      significant bytes (positions j+1..3), but it does not matter since we
137      already have a non-zero bit at position 8*j+7.
138 
139      So, the test whether any byte in longword1 is zero is equivalent to
140      testing whether tmp is nonzero.  */
141 
142   while (n >= sizeof (longword))
143     {
144       longword longword1 = *longword_ptr ^ repeated_c;
145 
146       if ((((longword1 - repeated_one) & ~longword1)
147            & (repeated_one << 7)) != 0)
148         break;
149       longword_ptr++;
150       n -= sizeof (longword);
151     }
152 
153   char_ptr = (const unsigned char *) longword_ptr;
154 
155   /* At this point, we know that either n < sizeof (longword), or one of the
156      sizeof (longword) bytes starting at char_ptr is == c.  On little-endian
157      machines, we could determine the first such byte without any further
158      memory accesses, just by looking at the tmp result from the last loop
159      iteration.  But this does not work on big-endian machines.  Choose code
160      that works in both cases.  */
161 
162   for (; n > 0; --n, ++char_ptr)
163     {
164       if (*char_ptr == c)
165         return (void *) char_ptr;
166     }
167 
168   return NULL;
169 }
170 #ifdef weak_alias
171 weak_alias (__memchr, BP_SYM (memchr))
172 #endif
173