string: Improve generic strnlen with memchr

It also cleanups the multiple inclusion by leaving the ifunc
implementation to undef the weak_alias and libc_hidden_def.

Co-authored-by: Richard Henderson  <richard.henderson@linaro.org>
Reviewed-by: Noah Goldstein <goldstein.w.n@gmail.com>

1 files changed, 5 insertions, 132 deletions

diff --git a/string/strnlen.c b/string/strnlen.c
index 6ff294e..dc23354 100644
--- a/string/strnlen.c
+++ b/string/strnlen.c
@@ -1,10 +1,6 @@
 /* Find the length of STRING, but scan at most MAXLEN characters.
    Copyright (C) 1991-2023 Free Software Foundation, Inc.
 
-   Based on strlen written by Torbjorn Granlund (tege@sics.se),
-   with help from Dan Sahlin (dan@sics.se);
-   commentary by Jim Blandy (jimb@ai.mit.edu).
-
    The GNU C Library 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
@@ -20,7 +16,6 @@
    not, see <https://www.gnu.org/licenses/>.  */
 
 #include <string.h>
-#include <stdlib.h>
 
 /* Find the length of S, but scan at most MAXLEN characters.  If no
    '\0' terminator is found in that many characters, return MAXLEN.  */
@@ -32,134 +27,12 @@
 size_t
 __strnlen (const char *str, size_t maxlen)
 {
-  const char *char_ptr, *end_ptr = str + maxlen;
-  const unsigned long int *longword_ptr;
-  unsigned long int longword, himagic, lomagic;
-
-  if (maxlen == 0)
-    return 0;
-
-  if (__glibc_unlikely (end_ptr < str))
-    end_ptr = (const char *) ~0UL;
-
-  /* Handle the first few characters by reading one character at a time.
-     Do this until CHAR_PTR is aligned on a longword boundary.  */
-  for (char_ptr = str; ((unsigned long int) char_ptr
-			& (sizeof (longword) - 1)) != 0;
-       ++char_ptr)
-    if (*char_ptr == '\0')
-      {
-	if (char_ptr > end_ptr)
-	  char_ptr = end_ptr;
-	return char_ptr - str;
-      }
-
-  /* All these elucidatory comments refer to 4-byte longwords,
-     but the theory applies equally well to 8-byte longwords.  */
-
-  longword_ptr = (unsigned long int *) char_ptr;
-
-  /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
-     the "holes."  Note that there is a hole just to the left of
-     each byte, with an extra at the end:
-
-     bits:  01111110 11111110 11111110 11111111
-     bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
-
-     The 1-bits make sure that carries propagate to the next 0-bit.
-     The 0-bits provide holes for carries to fall into.  */
-  himagic = 0x80808080L;
-  lomagic = 0x01010101L;
-  if (sizeof (longword) > 4)
-    {
-      /* 64-bit version of the magic.  */
-      /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
-      himagic = ((himagic << 16) << 16) | himagic;
-      lomagic = ((lomagic << 16) << 16) | lomagic;
-    }
-  if (sizeof (longword) > 8)
-    abort ();
-
-  /* Instead of the traditional loop which tests each character,
-     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 zero.  */
-  while (longword_ptr < (unsigned long int *) end_ptr)
-    {
-      /* We tentatively exit the loop if adding MAGIC_BITS to
-	 LONGWORD fails to change any of the hole bits of LONGWORD.
-
-	 1) Is this safe?  Will it catch all the zero bytes?
-	 Suppose there is a byte with all zeros.  Any carry bits
-	 propagating from its left will fall into the hole at its
-	 least significant bit and stop.  Since there will be no
-	 carry from its most significant bit, the LSB of the
-	 byte to the left will be unchanged, and the zero will be
-	 detected.
-
-	 2) Is this worthwhile?  Will it ignore everything except
-	 zero bytes?  Suppose every byte of LONGWORD has a bit set
-	 somewhere.  There will be a carry into bit 8.  If bit 8
-	 is set, this will carry into bit 16.  If bit 8 is clear,
-	 one of bits 9-15 must be set, so there will be a carry
-	 into bit 16.  Similarly, there will be a carry into bit
-	 24.  If one of bits 24-30 is set, there will be a carry
-	 into bit 31, so all of the hole bits will be changed.
-
-	 The one misfire occurs when bits 24-30 are clear and bit
-	 31 is set; in this case, the hole at bit 31 is not
-	 changed.  If we had access to the processor carry flag,
-	 we could close this loophole by putting the fourth hole
-	 at bit 32!
-
-	 So it ignores everything except 128's, when they're aligned
-	 properly.  */
-
-      longword = *longword_ptr++;
-
-      if ((longword - lomagic) & himagic)
-	{
-	  /* Which of the bytes was the zero?  If none of them were, it was
-	     a misfire; continue the search.  */
-
-	  const char *cp = (const char *) (longword_ptr - 1);
-
-	  char_ptr = cp;
-	  if (cp[0] == 0)
-	    break;
-	  char_ptr = cp + 1;
-	  if (cp[1] == 0)
-	    break;
-	  char_ptr = cp + 2;
-	  if (cp[2] == 0)
-	    break;
-	  char_ptr = cp + 3;
-	  if (cp[3] == 0)
-	    break;
-	  if (sizeof (longword) > 4)
-	    {
-	      char_ptr = cp + 4;
-	      if (cp[4] == 0)
-		break;
-	      char_ptr = cp + 5;
-	      if (cp[5] == 0)
-		break;
-	      char_ptr = cp + 6;
-	      if (cp[6] == 0)
-		break;
-	      char_ptr = cp + 7;
-	      if (cp[7] == 0)
-		break;
-	    }
-	}
-      char_ptr = end_ptr;
-    }
-
-  if (char_ptr > end_ptr)
-    char_ptr = end_ptr;
-  return char_ptr - str;
+  const char *found = memchr (str, '\0', maxlen);
+  return found ? found - str : maxlen;
 }
+
 #ifndef STRNLEN
-libc_hidden_def (__strnlen)
 weak_alias (__strnlen, strnlen)
-#endif
+libc_hidden_def (__strnlen)
 libc_hidden_def (strnlen)
+#endif