1 files changed, 0 insertions, 142 deletions
diff --git a/lib/strchrnul.c b/lib/strchrnul.c
deleted file mode 100644
index f6b07227..00000000
--- a/lib/strchrnul.c
+++ /dev/null
@@ -1,142 +0,0 @@
-/* Searching in a string.
-   Copyright (C) 2003, 2007-2013 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;
-}