lib/memchr.c

   1 /* Copyright (C) 1991 Free Software Foundation, Inc.
   2    Based on strlen implemention by Torbjorn Granlund (tege@sics.se),
   3    with help from Dan Sahlin (dan@sics.se) and
   4    commentary by Jim Blandy (jimb@ai.mit.edu);
   5    adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
   6    and implemented by Roland McGrath (roland@ai.mit.edu).
   7
   8 The GNU C Library is free software; you can redistribute it and/or
   9 modify it under the terms of the GNU Library General Public License as
  10 published by the Free Software Foundation; either version 2 of the
  11 License, or (at your option) any later version.
  12
  13 The GNU C Library is distributed in the hope that it will be useful,
  14 but WITHOUT ANY WARRANTY; without even the implied warranty of
  15 MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  16 Library General Public License for more details.
  17
  18 You should have received a copy of the GNU Library General Public
  19 License along with the GNU C Library; see the file COPYING.LIB.  If
  20 not, write to the Free Software Foundation, Inc., 675 Mass Ave,
  21 Cambridge, MA 02139, USA.  */
  22
  23
  24
  25
  26 /* Search no more than N bytes of S for C.  */
  27
  28 char *
  29 memchr(s, c, n)
  30      unsigned char * s ;
  31      int c ;
  32      unsigned n;
  33 {
  34   unsigned char *char_ptr;
  35   unsigned long int *longword_ptr;
  36   unsigned long int longword, magic_bits, charmask;
  37
  38   c = (unsigned char) c;
  39
  40   /* Handle the first few characters by reading one character at a time.
  41      Do this until CHAR_PTR is aligned on a 4-byte border.  */
  42   for (char_ptr = s; n > 0 && ((unsigned long int) char_ptr & 3) != 0;
  43        --n, ++char_ptr)
  44     if (*char_ptr == c)
  45       return (char *) char_ptr;
  46
  47   longword_ptr = (unsigned long int *) char_ptr;
  48
  49   /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
  50      the "holes."  Note that there is a hole just to the left of
  51      each byte, with an extra at the end:
  52
  53      bits:  01111110 11111110 11111110 11111111
  54      bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
  55
  56      The 1-bits make sure that carries propagate to the next 0-bit.
  57      The 0-bits provide holes for carries to fall into.  */
  58   magic_bits = 0x7efefeff;
  59
  60   /* Set up a longword, each of whose bytes is C.  */
  61   charmask = c | (c << 8);
  62   charmask |= charmask << 16;
  63
  64   /* Instead of the traditional loop which tests each character,
  65      we will test a longword at a time.  The tricky part is testing
  66      if *any of the four* bytes in the longword in question are zero.  */
  67   while (n >= 4)
  68     {
  69       /* We tentatively exit the loop if adding MAGIC_BITS to
  70          LONGWORD fails to change any of the hole bits of LONGWORD.
  71
  72          1) Is this safe?  Will it catch all the zero bytes?
  73          Suppose there is a byte with all zeros.  Any carry bits
  74          propagating from its left will fall into the hole at its
  75          least significant bit and stop.  Since there will be no
  76          carry from its most significant bit, the LSB of the
  77          byte to the left will be unchanged, and the zero will be
  78          detected.
  79
  80          2) Is this worthwhile?  Will it ignore everything except
  81          zero bytes?  Suppose every byte of LONGWORD has a bit set
  82          somewhere.  There will be a carry into bit 8.  If bit 8
  83          is set, this will carry into bit 16.  If bit 8 is clear,
  84          one of bits 9-15 must be set, so there will be a carry
  85          into bit 16.  Similarly, there will be a carry into bit
  86          24.  If one of bits 24-30 is set, there will be a carry
  87          into bit 31, so all of the hole bits will be changed.
  88
  89          The one misfire occurs when bits 24-30 are clear and bit
  90          31 is set; in this case, the hole at bit 31 is not
  91          changed.  If we had access to the processor carry flag,
  92          we could close this loophole by putting the fourth hole
  93          at bit 32!
  94
  95          So it ignores everything except 128's, when they're aligned
  96          properly.
  97
  98          3) But wait!  Aren't we looking for C, not zero?
  99          Good point.  So what we do is XOR LONGWORD with a longword,
 100          each of whose bytes is C.  This turns each byte that is C
 101          into a zero.  */
 102
 103       longword = *longword_ptr++ ^ charmask;
 104
 105       /* Add MAGIC_BITS to LONGWORD.  */
 106       if ((((longword + magic_bits)
 107
 108             /* Set those bits that were unchanged by the addition.  */
 109             ^ ~longword)
 110
 111            /* Look at only the hole bits.  If any of the hole bits
 112               are unchanged, most likely one of the bytes was a
 113               zero.  */
 114            & ~magic_bits) != 0)
 115         {
 116           /* Which of the bytes was C?  If none of them were, it was
 117              a misfire; continue the search.  */
 118
 119            unsigned char *cp = ( unsigned char *) (longword_ptr - 1);
 120
 121           if (cp[0] == c)
 122             return (char *) cp;
 123           if (cp[1] == c)
 124             return (char *) &cp[1];
 125           if (cp[2] == c)
 126             return (char *) &cp[2];
 127           if (cp[3] == c)
 128             return (char *) &cp[3];
 129         }
 130
 131       n -= 4;
 132     }
 133
 134   char_ptr = ( unsigned char *) longword_ptr;
 135
 136   while (n-- > 0)
 137     {
 138       if (*char_ptr == c)
 139         return (char *) char_ptr;
 140       else
 141         ++char_ptr;
 142     }
 143
 144   return 0;
 145 }