2.5-18.1

1 files changed, 161 insertions, 0 deletions

diff --git a/string/strnlen.c b/string/strnlen.c
new file mode 100644
index 0000000..454257b
--- /dev/null
+++ b/string/strnlen.c
@@ -0,0 +1,161 @@
+/* Find the length of STRING, but scan at most MAXLEN characters.
+   Copyright (C) 1991,1993,1997,2000,2001,2005 Free Software Foundation, Inc.
+   Contributed by Jakub Jelinek <jakub@redhat.com>.
+
+   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
+   License, or (at your option) any later version.
+
+   The GNU C Library 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 the GNU C Library; see the file COPYING.LIB.  If not,
+   write to the Free Software Foundation, Inc., 59 Temple Place - Suite 330,
+   Boston, MA 02111-1307, USA.  */
+
+#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.  */
+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, magic_bits, himagic, lomagic;
+
+  if (maxlen == 0)
+    return 0;
+
+  if (__builtin_expect (end_ptr < str, 0))
+    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.  */
+  magic_bits = 0x7efefeffL;
+  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.  */
+      magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
+      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;
+}
+weak_alias (__strnlen, strnlen)
+libc_hidden_def (strnlen)