--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/lib/strchrnul.c	Fri Oct 29 19:20:39 2010 -0600
@@ -0,0 +1,142 @@
+/* Searching in a string.
+   Copyright (C) 2003, 2007, 2008, 2009, 2010 Free Software Foundation, Inc.
+
+   This program is free software: you can redistribute it and/or modify
+   it under the terms of the GNU General Public License as published by
+   the Free Software Foundation; either version 3 of the License, or
+   (at your option) any later version.
+
+   This program 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 General Public License for more details.
+
+   You should have received a copy of the GNU General Public License
+   along with this program.  If not, see <http://www.gnu.org/licenses/>.  */
+
+#include <config.h>
+
+/* Specification.  */
+#include <string.h>
+
+/* Find the first occurrence of C in S or the final NUL byte.  */
+char *
+strchrnul (const char *s, int c_in)
+{
+  /* On 32-bit hardware, choosing longword to be a 32-bit unsigned
+     long instead of a 64-bit uintmax_t tends to give better
+     performance.  On 64-bit hardware, unsigned long is generally 64
+     bits already.  Change this typedef to experiment with
+     performance.  */
+  typedef unsigned long int longword;
+
+  const unsigned char *char_ptr;
+  const longword *longword_ptr;
+  longword repeated_one;
+  longword repeated_c;
+  unsigned char c;
+
+  c = (unsigned char) c_in;
+  if (!c)
+    return rawmemchr (s, 0);
+
+  /* Handle the first few bytes by reading one byte at a time.
+     Do this until CHAR_PTR is aligned on a longword boundary.  */
+  for (char_ptr = (const unsigned char *) s;
+       (size_t) char_ptr % sizeof (longword) != 0;
+       ++char_ptr)
+    if (!*char_ptr || *char_ptr == c)
+      return (char *) char_ptr;
+
+  longword_ptr = (const longword *) char_ptr;
+
+  /* All these elucidatory comments refer to 4-byte longwords,
+     but the theory applies equally well to any size longwords.  */
+
+  /* Compute auxiliary longword values:
+     repeated_one is a value which has a 1 in every byte.
+     repeated_c has c in every byte.  */
+  repeated_one = 0x01010101;
+  repeated_c = c | (c << 8);
+  repeated_c |= repeated_c << 16;
+  if (0xffffffffU < (longword) -1)
+    {
+      repeated_one |= repeated_one << 31 << 1;
+      repeated_c |= repeated_c << 31 << 1;
+      if (8 < sizeof (longword))
+        {
+          size_t i;
+
+          for (i = 64; i < sizeof (longword) * 8; i *= 2)
+            {
+              repeated_one |= repeated_one << i;
+              repeated_c |= repeated_c << i;
+            }
+        }
+    }
+
+  /* Instead of the traditional loop which tests each byte, 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 equal to NUL or
+     c.  We first use an xor with repeated_c.  This reduces the task
+     to testing whether *any of the four* bytes in longword1 or
+     longword2 is zero.
+
+     Let's consider longword1.  We compute tmp =
+       ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7).
+     That is, we perform the following operations:
+       1. Subtract repeated_one.
+       2. & ~longword1.
+       3. & a mask consisting of 0x80 in every byte.
+     Consider what happens in each byte:
+       - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff,
+         and step 3 transforms it into 0x80.  A carry can also be propagated
+         to more significant bytes.
+       - If a byte of longword1 is nonzero, let its lowest 1 bit be at
+         position k (0 <= k <= 7); so the lowest k bits are 0.  After step 1,
+         the byte ends in a single bit of value 0 and k bits of value 1.
+         After step 2, the result is just k bits of value 1: 2^k - 1.  After
+         step 3, the result is 0.  And no carry is produced.
+     So, if longword1 has only non-zero bytes, tmp is zero.
+     Whereas if longword1 has a zero byte, call j the position of the least
+     significant zero byte.  Then the result has a zero at positions 0, ...,
+     j-1 and a 0x80 at position j.  We cannot predict the result at the more
+     significant bytes (positions j+1..3), but it does not matter since we
+     already have a non-zero bit at position 8*j+7.
+
+     The test whether any byte in longword1 or longword2 is zero is equivalent
+     to testing whether tmp1 is nonzero or tmp2 is nonzero.  We can combine
+     this into a single test, whether (tmp1 | tmp2) is nonzero.
+
+     This test can read more than one byte beyond the end of a string,
+     depending on where the terminating NUL is encountered.  However,
+     this is considered safe since the initialization phase ensured
+     that the read will be aligned, therefore, the read will not cross
+     page boundaries and will not cause a fault.  */
+
+  while (1)
+    {
+      longword longword1 = *longword_ptr ^ repeated_c;
+      longword longword2 = *longword_ptr;
+
+      if (((((longword1 - repeated_one) & ~longword1)
+            | ((longword2 - repeated_one) & ~longword2))
+           & (repeated_one << 7)) != 0)
+        break;
+      longword_ptr++;
+    }
+
+  char_ptr = (const unsigned char *) longword_ptr;
+
+  /* At this point, we know that one of the sizeof (longword) bytes
+     starting at char_ptr is == 0 or == c.  On little-endian machines,
+     we could determine the first such byte without any further memory
+     accesses, just by looking at the tmp result from the last loop
+     iteration.  But this does not work on big-endian machines.
+     Choose code that works in both cases.  */
+
+  char_ptr = (unsigned char *) longword_ptr;
+  while (*char_ptr && (*char_ptr != c))
+    char_ptr++;
+  return (char *) char_ptr;
+}