Update/gnulib 2026 03 (#2247)HEADmaster

* Sync with the 202601-stable Gnulib code (4a3650d887)

* Ignore more deps stuff in gnulib

* Remove autogenerated gnulib files

* Ignore more gnulib generated headers

1 files changed, 86 insertions, 88 deletions

diff --git a/gl/memchr.c b/gl/memchr.c
index ef0d15f7..6adac7e1 100644
--- a/gl/memchr.c
+++ b/gl/memchr.c
@@ -1,4 +1,4 @@
-/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2025
+/* Copyright (C) 1991, 1993, 1996-1997, 1999-2000, 2003-2004, 2006, 2008-2026
    Free Software Foundation, Inc.
 
    Based on strlen implementation by Torbjorn Granlund (tege@sics.se),
@@ -65,105 +65,103 @@ __memchr (void const *s, int c_in, size_t n)
      performance.  */
   typedef unsigned long int longword;
 
-  const unsigned char *char_ptr;
-  const longword *longword_ptr;
-  longword repeated_one;
-  longword repeated_c;
-  unsigned reg_char c;
+  unsigned reg_char c = (unsigned char) c_in;
 
-  c = (unsigned char) c_in;
+  const longword *longword_ptr;
 
   /* 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;
-       n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
-       --n, ++char_ptr)
-    if (*char_ptr == c)
-      return (void *) char_ptr;
+  {
+    const unsigned char *char_ptr;
+    for (char_ptr = (const unsigned char *) s;
+         n > 0 && (size_t) char_ptr % sizeof (longword) != 0;
+         --n, ++char_ptr)
+      if (*char_ptr == c)
+        return (void *) char_ptr;
 
-  longword_ptr = (const longword *) 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)
+  {
+    /* 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 = 0x01010101;
+    longword 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))
+          for (size_t 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 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.
-
-     So, the test whether any byte in longword1 is zero is equivalent to
-     testing whether tmp is nonzero.  */
-
-  while (n >= sizeof (longword))
-    {
-      longword longword1 = *longword_ptr ^ repeated_c;
-
-      if ((((longword1 - repeated_one) & ~longword1)
-           & (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 == 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.  */
-
-  for (; n > 0; --n, ++char_ptr)
-    {
-      if (*char_ptr == c)
-        return (void *) char_ptr;
-    }
+      }
+
+    /* 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 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.
+
+       So, the test whether any byte in longword1 is zero is equivalent to
+       testing whether tmp is nonzero.  */
+
+    while (n >= sizeof (longword))
+      {
+        longword longword1 = *longword_ptr ^ repeated_c;
+
+        if ((((longword1 - repeated_one) & ~longword1)
+             & (repeated_one << 7)) != 0)
+          break;
+        longword_ptr++;
+        n -= sizeof (longword);
+      }
+  }
+
+  {
+    const unsigned char *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 == 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.  */
+
+    for (; n > 0; --n, ++char_ptr)
+      {
+        if (*char_ptr == c)
+          return (void *) char_ptr;
+      }
+  }
 
   return NULL;
 }