text: /* Copyright (C) 1991, 1993 Free Software Foundation, Inc.
     Based on strlen implemention by Torbjorn Granlund (tege@sics.se),
     with help from Dan Sahlin (dan@sics.se) and
     commentary by Jim Blandy (jimb@ai.mit.edu);
     adaptation to memchr suggested by Dick Karpinski (dick@cca.ucsf.edu),
     and implemented by Roland McGrath (roland@ai.mit.edu).
  
  The GNU C Library is free software; you can redistribute it and/or
  modify it under the terms of the GNU Library General Public License as
  published by the Free Software Foundation; either version 2 of the
  License, or (at your option) any later version.
  
  The GNU C Library is distributed in the hope that it will be useful,
  but WITHOUT ANY WARRANTY; without even the implied warranty of
  MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
  Library General Public License for more details.
  
  You should have received a copy of the GNU Library General Public
  License along with the GNU C Library; see the file COPYING.LIB.  If
  not, write to the Free Software Foundation, Inc., 675 Mass Ave,
  Cambridge, MA 02139, USA.  */
  
  #include <ansidecl.h>
  #include <string.h>
  
  
  /* Search no more than N bytes of S for C.  */
  
  PTR
  DEFUN(memchr, (s, c, n), CONST PTR s AND int c AND size_t n)
  {
    CONST unsigned char *char_ptr;
    CONST unsigned long int *longword_ptr;
    unsigned long int longword, magic_bits, charmask;
  
    c = (unsigned char) c;
  
    /* Handle the first few characters by reading one character at a time.
       Do this until CHAR_PTR is aligned on a longword boundary.  */
    for (char_ptr = s; n > 0 && ((unsigned long int) char_ptr
  			       & (sizeof (longword) - 1)) != 0;
         --n, ++char_ptr)
      if (*char_ptr == c)
        return (PTR) char_ptr;
  
    /* All these elucidatory comments refer to 4-byte longwords,
       but the theory applies equally well to 8-byte longwords.  */
  
    longword_ptr = (unsigned long int *) char_ptr;
  
    /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
       the "holes."  Note that there is a hole just to the left of
       each byte, with an extra at the end:
       
       bits:  01111110 11111110 11111110 11111111
       bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD 
  
       The 1-bits make sure that carries propagate to the next 0-bit.
       The 0-bits provide holes for carries to fall into.  */
    switch (sizeof (longword))
      {
      case 4: magic_bits = 0x7efefeffL; break;
      case 8: magic_bits = (0x7efefefeL << 32) | 0xfefefeffL; break;
      default:
        abort ();
      }
  
    /* Set up a longword, each of whose bytes is C.  */
    charmask = c | (c << 8);
    charmask |= charmask << 16;
    if (sizeof (longword) > 4)
      charmask |= charmask << 32;
    if (sizeof (longword) > 8)
      abort ();
  
    /* Instead of the traditional loop which tests each character,
       we will test a longword at a time.  The tricky part is testing
       if *any of the four* bytes in the longword in question are zero.  */
    while (n >= sizeof (longword))
      {
        /* We tentatively exit the loop if adding MAGIC_BITS to
  	 LONGWORD fails to change any of the hole bits of LONGWORD.
  
  	 1) Is this safe?  Will it catch all the zero bytes?
  	 Suppose there is a byte with all zeros.  Any carry bits
  	 propagating from its left will fall into the hole at its
  	 least significant bit and stop.  Since there will be no
  	 carry from its most significant bit, the LSB of the
  	 byte to the left will be unchanged, and the zero will be
  	 detected.
  
  	 2) Is this worthwhile?  Will it ignore everything except
  	 zero bytes?  Suppose every byte of LONGWORD has a bit set
  	 somewhere.  There will be a carry into bit 8.  If bit 8
  	 is set, this will carry into bit 16.  If bit 8 is clear,
  	 one of bits 9-15 must be set, so there will be a carry
  	 into bit 16.  Similarly, there will be a carry into bit
  	 24.  If one of bits 24-30 is set, there will be a carry
  	 into bit 31, so all of the hole bits will be changed.
  
  	 The one misfire occurs when bits 24-30 are clear and bit
  	 31 is set; in this case, the hole at bit 31 is not
  	 changed.  If we had access to the processor carry flag,
  	 we could close this loophole by putting the fourth hole
  	 at bit 32!
  
  	 So it ignores everything except 128's, when they're aligned
  	 properly.
  
  	 3) But wait!  Aren't we looking for C, not zero?
  	 Good point.  So what we do is XOR LONGWORD with a longword,
  	 each of whose bytes is C.  This turns each byte that is C
  	 into a zero.  */
  
        longword = *longword_ptr++ ^ charmask;
  
        /* Add MAGIC_BITS to LONGWORD.  */
        if ((((longword + magic_bits)
  	
  	    /* Set those bits that were unchanged by the addition.  */
  	    ^ ~longword)
  	       
  	   /* Look at only the hole bits.  If any of the hole bits
  	      are unchanged, most likely one of the bytes was a
  	      zero.  */
  	   & ~magic_bits) != 0)
  	{
  	  /* Which of the bytes was C?  If none of them were, it was
  	     a misfire; continue the search.  */
  
  	  CONST unsigned char *cp = (CONST unsigned char *) (longword_ptr - 1);
  
  	  if (cp[0] == c)
  	    return (PTR) cp;
  	  if (cp[1] == c)
  	    return (PTR) &cp[1];
  	  if (cp[2] == c)
  	    return (PTR) &cp[2];
  	  if (cp[3] == c)
  	    return (PTR) &cp[3];
  	  if (sizeof (longword) > 4)
  	    {
  	      if (cp[4] == c)
  		return (PTR) &cp[4];
  	      if (cp[5] == c)
  		return (PTR) &cp[5];
  	      if (cp[6] == c)
  		return (PTR) &cp[6];
  	      if (cp[7] == c)
  		return (PTR) &cp[7];
  	    }
  	}
  
        n -= sizeof (longword);
      }
  
    char_ptr = (CONST unsigned char *) longword_ptr;
  
    while (n-- > 0)
      {
        if (*char_ptr == c)
  	return (PTR) char_ptr;
        else
  	++char_ptr;
      }
  
    return NULL;
  }
  