1 files changed, 125 insertions, 0 deletions
diff --git a/gl/lib/rawmemchr.c b/gl/lib/rawmemchr.c
new file mode 100644
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--- /dev/null
+++ b/gl/lib/rawmemchr.c
@@ -0,0 +1,125 @@
+/* Searching in a string.
+   Copyright (C) 2008-2022 Free Software Foundation, Inc.
+
+   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>
+
+/* Specification.  */
+#include <string.h>
+
+/* A function definition is only needed if HAVE_RAWMEMCHR is not defined.  */
+#if !HAVE_RAWMEMCHR
+
+# include <limits.h>
+# include <stdalign.h>
+# include <stdint.h>
+
+# include "verify.h"
+
+/* Find the first occurrence of C in S.  */
+void *
+rawmemchr (const void *s, int c_in)
+{
+  /* Change this typedef to experiment with performance.  */
+  typedef uintptr_t longword;
+  /* If you change the "uintptr_t", you should change UINTPTR_WIDTH to match.
+     This verifies that the type does not have padding bits.  */
+  verify (UINTPTR_WIDTH == UCHAR_WIDTH * sizeof (longword));
+
+  const unsigned char *char_ptr;
+  unsigned char c = 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;
+       (uintptr_t) char_ptr % alignof (longword) != 0;
+       ++char_ptr)
+    if (*char_ptr == c)
+      return (void *) char_ptr;
+
+  longword const *longword_ptr = s = char_ptr;
+
+  /* Compute auxiliary longword values:
+     repeated_one is a value which has a 1 in every byte.
+     repeated_c has c in every byte.  */
+  longword repeated_one = (longword) -1 / UCHAR_MAX;
+  longword repeated_c = repeated_one * c;
+  longword repeated_hibit = repeated_one * (UCHAR_MAX / 2 + 1);
+
+  /* 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 bytes in the longword in question are equal to
+     c.  We first use an xor with repeated_c.  This reduces the task
+     to testing whether any of the bytes in longword1 is zero.
+
+     (The following comments assume 8-bit bytes, as POSIX requires;
+     the code's use of UCHAR_MAX should work even if bytes have more
+     than 8 bits.)
+
+     We compute tmp =
+       ((longword1 - repeated_one) & ~longword1) & (repeated_one * 0x80).
+     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_hibit) != 0)
+        break;
+      longword_ptr++;
+    }
+
+  char_ptr = s = longword_ptr;
+
+  /* At this point, we know that one of the sizeof (longword) bytes
+     starting at char_ptr is == c.  If we knew endianness, we
+     could determine the first such byte without any further memory
+     accesses, just by looking at the tmp result from the last loop
+     iteration.  However, the following simple and portable code does
+     not attempt this potential optimization.  */
+
+  while (*char_ptr != c)
+    char_ptr++;
+  return (void *) char_ptr;
+}
+
+#endif