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      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 (at) sics.se),
      5    with help from Dan Sahlin (dan (at) sics.se) and
      6    commentary by Jim Blandy (jimb (at) ai.mit.edu);
      7    adaptation to memchr suggested by Dick Karpinski (dick (at) cca.ucsf.edu),
      8    and implemented by Roland McGrath (roland (at) 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 (at) 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
    173