Original patch from: gentoo/src/patchsets/glibc/2.9/1020_all_glibc-2.9-strlen-hack.patch

-= BEGIN original header =-
http://sourceware.org/bugzilla/show_bug.cgi?id=5807
http://www.cl.cam.ac.uk/~am21/progtricks.html

-= END original header =-

diff -durN glibc-2_9.orig/string/strlen.c glibc-2_9/string/strlen.c
--- glibc-2_9.orig/string/strlen.c	2005-12-14 12:09:07.000000000 +0100
+++ glibc-2_9/string/strlen.c	2009-02-02 22:00:51.000000000 +0100
@@ -32,7 +32,7 @@
 {
   const char *char_ptr;
   const unsigned long int *longword_ptr;
-  unsigned long int longword, magic_bits, himagic, lomagic;
+  unsigned long int longword, himagic, lomagic;
 
   /* Handle the first few characters by reading one character at a time.
      Do this until CHAR_PTR is aligned on a longword boundary.  */
@@ -42,28 +42,14 @@
     if (*char_ptr == '\0')
       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;
     }
@@ -75,56 +61,12 @@
      if *any of the four* bytes in the longword in question are zero.  */
   for (;;)
     {
-      /* 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 (
-#if 0
-	  /* Add MAGIC_BITS to LONGWORD.  */
-	  (((longword + magic_bits)
-
-	    /* Set those bits that were unchanged by the addition.  */
-	    ^ ~longword)
-
-	   /* Look at only the hole bits.  If any of the hole bits
-	      are unchanged, most likely one of the bytes was a
-	      zero.  */
-	   & ~magic_bits)
-#else
-	  ((longword - lomagic) & himagic)
-#endif
-	  != 0)
+      /* This hack taken from Alan Mycroft's HAKMEMC postings.
+         See: http://www.cl.cam.ac.uk/~am21/progtricks.html */
+      if (((longword - lomagic) & ~longword & himagic) != 0)
 	{
-	  /* 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);