lib/memrchr.c

   1 /* memrchr -- find the last occurrence of a byte in a memory block
   2
   3    Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2015 Free Software
   4    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 3 of the License, or
  15    (at your option) 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
  23    along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
  24
  25 #if defined _LIBC
  26 # include <memcopy.h>
  27 #else
  28 # include <config.h>
  29 # define reg_char char
  30 #endif
  31
  32 #include <string.h>
  33 #include <limits.h>
  34
  35 #undef __memrchr
  36 #ifdef _LIBC
  37 # undef memrchr
  38 #endif
  39
  40 #ifndef weak_alias
  41 # define __memrchr memrchr
  42 #endif
  43
  44 /* Search no more than N bytes of S for C.  */
  45 void *
  46 __memrchr (void const *s, int c_in, size_t n)
  47 {
  48   /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
  49      long instead of a 64-bit uintmax_t tends to give better
  50      performance.  On 64-bit hardware, unsigned long is generally 64
  51      bits already.  Change this typedef to experiment with
  52      performance.  */
  53   typedef unsigned long int longword;
  54
  55   const unsigned char *char_ptr;
  56   const longword *longword_ptr;
  57   longword repeated_one;
  58   longword repeated_c;
  59   unsigned reg_char c;
  60
  61   c = (unsigned char) c_in;
  62
  63   /* Handle the last few bytes by reading one byte at a time.
  64      Do this until CHAR_PTR is aligned on a longword boundary.  */
  65   for (char_ptr = (const unsigned char *) s + n;
  66        n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
  67        --n)
  68     if (*--char_ptr == c)
  69       return (void *) char_ptr;
  70
  71   longword_ptr = (const longword *) char_ptr;
  72
  73   /* All these elucidatory comments refer to 4-byte longwords,
  74      but the theory applies equally well to any size longwords.  */
  75
  76   /* Compute auxiliary longword values:
  77      repeated_one is a value which has a 1 in every byte.
  78      repeated_c has c in every byte.  */
  79   repeated_one = 0x01010101;
  80   repeated_c = c | (c << 8);
  81   repeated_c |= repeated_c << 16;
  82   if (0xffffffffU < (longword) -1)
  83     {
  84       repeated_one |= repeated_one << 31 << 1;
  85       repeated_c |= repeated_c << 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_c |= repeated_c << i;
  94             }
  95         }
  96     }
  97
  98   /* Instead of the traditional loop which tests each byte, we will test a
  99      longword at a time.  The tricky part is testing if *any of the four*
 100      bytes in the longword in question are equal to c.  We first use an xor
 101      with repeated_c.  This reduces the task to testing whether *any of the
 102      four* bytes in longword1 is zero.
 103
 104      We compute tmp =
 105        ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
 106      That is, we perform the following operations:
 107        1. Subtract repeated_one.
 108        2. & ~longword1.
 109        3. & a mask consisting of 0x80 in every byte.
 110      Consider what happens in each byte:
 111        - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
 112          and step 3 transforms it into 0x80.  A carry can also be propagated
 113          to more significant bytes.
 114        - If a byte of longword1 is nonzero, let its lowest 1 bit be at
 115          position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
 116          the byte ends in a single bit of value 0 and k bits of value 1.
 117          After step 2, the result is just k bits of value 1: 2^k - 1.  After
 118          step 3, the result is 0.  And no carry is produced.
 119      So, if longword1 has only non-zero bytes, tmp is zero.
 120      Whereas if longword1 has a zero byte, call j the position of the least
 121      significant zero byte.  Then the result has a zero at positions 0, ...,
 122      j-1 and a 0x80 at position j.  We cannot predict the result at the more
 123      significant bytes (positions j+1..3), but it does not matter since we
 124      already have a non-zero bit at position 8*j+7.
 125
 126      So, the test whether any byte in longword1 is zero is equivalent to
 127      testing whether tmp is nonzero.  */
 128
 129   while (n >= sizeof (longword))
 130     {
 131       longword longword1 = *--longword_ptr ^ repeated_c;
 132
 133       if ((((longword1 - repeated_one) & ~longword1)
 134            & (repeated_one << 7)) != 0)
 135         {
 136           longword_ptr++;
 137           break;
 138         }
 139       n -= sizeof (longword);
 140     }
 141
 142   char_ptr = (const unsigned char *) longword_ptr;
 143
 144   /* At this point, we know that either n < sizeof (longword), or one of the
 145      sizeof (longword) bytes starting at char_ptr is == c.  On little-endian
 146      machines, we could determine the first such byte without any further
 147      memory accesses, just by looking at the tmp result from the last loop
 148      iteration.  But this does not work on big-endian machines.  Choose code
 149      that works in both cases.  */
 150
 151   while (n-- > 0)
 152     {
 153       if (*--char_ptr == c)
 154         return (void *) char_ptr;
 155     }
 156
 157   return NULL;
 158 }
 159 #ifdef weak_alias
 160 weak_alias (__memrchr, memrchr)
 161 #endif