gl/memchr.c

   1 /* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2012
   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 <http://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 *
  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