Restoring authorship annotation for <axc@yandex-team.ru>. Commit 1 of 2.

1 files changed, 136 insertions, 136 deletions

diff --git a/contrib/tools/bison/gnulib/src/rawmemchr.c b/contrib/tools/bison/gnulib/src/rawmemchr.c
index a0298ce64e..b0f4f4d044 100644
--- a/contrib/tools/bison/gnulib/src/rawmemchr.c
+++ b/contrib/tools/bison/gnulib/src/rawmemchr.c
@@ -1,136 +1,136 @@
-/* Searching in a string.
-   Copyright (C) 2008-2013 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.  */
-void *
-rawmemchr (const void *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;
-
-  /* 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 == c)
-      return (void *) 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 is zero.
-
-     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 is zero is equivalent
-     to testing whether tmp is nonzero.
-
-     This test can read beyond the end of a string, depending on where
-     C_IN 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;
-
-      if ((((longword1 - repeated_one) & ~longword1)
-           & (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 == 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 != c)
-    char_ptr++;
-  return (void *) char_ptr;
-}
+/* Searching in a string. 
+   Copyright (C) 2008-2013 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.  */ 
+void * 
+rawmemchr (const void *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; 
+ 
+  /* 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 == c) 
+      return (void *) 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 is zero. 
+ 
+     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 is zero is equivalent 
+     to testing whether tmp is nonzero. 
+ 
+     This test can read beyond the end of a string, depending on where 
+     C_IN 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; 
+ 
+      if ((((longword1 - repeated_one) & ~longword1) 
+           & (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 == 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 != c) 
+    char_ptr++; 
+  return (void *) char_ptr; 
+}