Adding upstream version 9.1.upstream/9.1upstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 169 insertions, 0 deletions

diff --git a/lib/memchr2.c b/lib/memchr2.c
new file mode 100644
index 0000000..ef833df
--- /dev/null
+++ b/lib/memchr2.c
@@ -0,0 +1,169 @@
+/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2022
+   Free Software Foundation, Inc.
+
+   Based on strlen implementation 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 in glibc by Roland McGrath (roland@ai.mit.edu).
+   Extension to memchr2 implemented by Eric Blake (ebb9@byu.net).
+
+   This file is free software: you can redistribute it and/or modify
+   it under the terms of the GNU Lesser General Public License as
+   published by the Free Software Foundation; either version 2.1 of the
+   License, or (at your option) any later version.
+
+   This file 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 Lesser General Public License for more details.
+
+   You should have received a copy of the GNU Lesser General Public License
+   along with this program.  If not, see <https://www.gnu.org/licenses/>.  */
+
+#include <config.h>
+
+#include "memchr2.h"
+
+#include <limits.h>
+#include <stdint.h>
+#include <string.h>
+
+/* Return the first address of either C1 or C2 (treated as unsigned
+   char) that occurs within N bytes of the memory region S.  If
+   neither byte appears, return NULL.  */
+void *
+memchr2 (void const *s, int c1_in, int c2_in, size_t n)
+{
+  /* 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;
+  void const *void_ptr;
+  const longword *longword_ptr;
+  longword repeated_one;
+  longword repeated_c1;
+  longword repeated_c2;
+  unsigned char c1;
+  unsigned char c2;
+
+  c1 = (unsigned char) c1_in;
+  c2 = (unsigned char) c2_in;
+
+  if (c1 == c2)
+    return memchr (s, c1, n);
+
+  /* Handle the first few bytes by reading one byte at a time.
+     Do this until VOID_PTR is aligned on a longword boundary.  */
+  for (void_ptr = s;
+       n > 0 && (uintptr_t) void_ptr % sizeof (longword) != 0;
+       --n)
+    {
+      char_ptr = void_ptr;
+      if (*char_ptr == c1 || *char_ptr == c2)
+        return (void *) void_ptr;
+      void_ptr = char_ptr + 1;
+    }
+
+  longword_ptr = void_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_c1 has c1 in every byte.
+     repeated_c2 has c2 in every byte.  */
+  repeated_one = 0x01010101;
+  repeated_c1 = c1 | (c1 << 8);
+  repeated_c2 = c2 | (c2 << 8);
+  repeated_c1 |= repeated_c1 << 16;
+  repeated_c2 |= repeated_c2 << 16;
+  if (0xffffffffU < (longword) -1)
+    {
+      repeated_one |= repeated_one << 31 << 1;
+      repeated_c1 |= repeated_c1 << 31 << 1;
+      repeated_c2 |= repeated_c2 << 31 << 1;
+      if (8 < sizeof (longword))
+        {
+          size_t i;
+
+          for (i = 64; i < sizeof (longword) * 8; i *= 2)
+            {
+              repeated_one |= repeated_one << i;
+              repeated_c1 |= repeated_c1 << i;
+              repeated_c2 |= repeated_c2 << 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 c1 or c2.  We first use
+     an xor with repeated_c1 and repeated_c2, respectively.  This reduces
+     the task to testing whether *any of the four* bytes in longword1 or
+     longword2 is zero.
+
+     Let's consider longword1.  We compute tmp1 =
+       ((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, tmp1 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.
+
+     Similarly, we compute tmp2 =
+       ((longword2 - repeated_one) & ~longword2) & (repeated_one << 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.  */
+
+  while (n >= sizeof (longword))
+    {
+      longword longword1 = *longword_ptr ^ repeated_c1;
+      longword longword2 = *longword_ptr ^ repeated_c2;
+
+      if (((((longword1 - repeated_one) & ~longword1)
+            | ((longword2 - repeated_one) & ~longword2))
+           & (repeated_one << 7)) != 0)
+        break;
+      longword_ptr++;
+      n -= sizeof (longword);
+    }
+
+  char_ptr = (const unsigned char *) longword_ptr;
+
+  /* At this point, we know that either n < sizeof (longword), or one of the
+     sizeof (longword) bytes starting at char_ptr is == c1 or == c2.  On
+     little-endian machines, we could determine the first such byte without
+     any further memory accesses, just by looking at the (tmp1 | tmp2) result
+     from the last loop iteration.  But this does not work on big-endian
+     machines.  Choose code that works in both cases.  */
+
+  for (; n > 0; --n, ++char_ptr)
+    {
+      if (*char_ptr == c1 || *char_ptr == c2)
+        return (void *) char_ptr;
+    }
+
+  return NULL;
+}