1 files changed, 168 insertions, 0 deletions
diff --git a/sysdeps/generic/memchr.c b/sysdeps/generic/memchr.c
new file mode 100644
index 0000000000..d17f9c649e
--- /dev/null
+++ b/sysdeps/generic/memchr.c
@@ -0,0 +1,168 @@
+/* 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;
+}