patches/glibc/2.9/210-2.9-strlen-hack.patch
branchnewlib
changeset 1365 c4d124ed9f8e
     1.1 --- /dev/null	Thu Jan 01 00:00:00 1970 +0000
     1.2 +++ b/patches/glibc/2.9/210-2.9-strlen-hack.patch	Sun Apr 19 16:17:11 2009 +0000
     1.3 @@ -0,0 +1,109 @@
     1.4 +Original patch from: gentoo/src/patchsets/glibc/2.9/1020_all_glibc-2.9-strlen-hack.patch
     1.5 +
     1.6 +-= BEGIN original header =-
     1.7 +http://sourceware.org/bugzilla/show_bug.cgi?id=5807
     1.8 +http://www.cl.cam.ac.uk/~am21/progtricks.html
     1.9 +
    1.10 +-= END original header =-
    1.11 +
    1.12 +diff -durN glibc-2_9.orig/string/strlen.c glibc-2_9/string/strlen.c
    1.13 +--- glibc-2_9.orig/string/strlen.c	2005-12-14 12:09:07.000000000 +0100
    1.14 ++++ glibc-2_9/string/strlen.c	2009-02-02 22:00:51.000000000 +0100
    1.15 +@@ -32,7 +32,7 @@
    1.16 + {
    1.17 +   const char *char_ptr;
    1.18 +   const unsigned long int *longword_ptr;
    1.19 +-  unsigned long int longword, magic_bits, himagic, lomagic;
    1.20 ++  unsigned long int longword, himagic, lomagic;
    1.21 + 
    1.22 +   /* Handle the first few characters by reading one character at a time.
    1.23 +      Do this until CHAR_PTR is aligned on a longword boundary.  */
    1.24 +@@ -42,28 +42,14 @@
    1.25 +     if (*char_ptr == '\0')
    1.26 +       return char_ptr - str;
    1.27 + 
    1.28 +-  /* All these elucidatory comments refer to 4-byte longwords,
    1.29 +-     but the theory applies equally well to 8-byte longwords.  */
    1.30 +-
    1.31 +   longword_ptr = (unsigned long int *) char_ptr;
    1.32 + 
    1.33 +-  /* Bits 31, 24, 16, and 8 of this number are zero.  Call these bits
    1.34 +-     the "holes."  Note that there is a hole just to the left of
    1.35 +-     each byte, with an extra at the end:
    1.36 +-
    1.37 +-     bits:  01111110 11111110 11111110 11111111
    1.38 +-     bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
    1.39 +-
    1.40 +-     The 1-bits make sure that carries propagate to the next 0-bit.
    1.41 +-     The 0-bits provide holes for carries to fall into.  */
    1.42 +-  magic_bits = 0x7efefeffL;
    1.43 +   himagic = 0x80808080L;
    1.44 +   lomagic = 0x01010101L;
    1.45 +   if (sizeof (longword) > 4)
    1.46 +     {
    1.47 +       /* 64-bit version of the magic.  */
    1.48 +       /* Do the shift in two steps to avoid a warning if long has 32 bits.  */
    1.49 +-      magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
    1.50 +       himagic = ((himagic << 16) << 16) | himagic;
    1.51 +       lomagic = ((lomagic << 16) << 16) | lomagic;
    1.52 +     }
    1.53 +@@ -75,56 +61,12 @@
    1.54 +      if *any of the four* bytes in the longword in question are zero.  */
    1.55 +   for (;;)
    1.56 +     {
    1.57 +-      /* We tentatively exit the loop if adding MAGIC_BITS to
    1.58 +-	 LONGWORD fails to change any of the hole bits of LONGWORD.
    1.59 +-
    1.60 +-	 1) Is this safe?  Will it catch all the zero bytes?
    1.61 +-	 Suppose there is a byte with all zeros.  Any carry bits
    1.62 +-	 propagating from its left will fall into the hole at its
    1.63 +-	 least significant bit and stop.  Since there will be no
    1.64 +-	 carry from its most significant bit, the LSB of the
    1.65 +-	 byte to the left will be unchanged, and the zero will be
    1.66 +-	 detected.
    1.67 +-
    1.68 +-	 2) Is this worthwhile?  Will it ignore everything except
    1.69 +-	 zero bytes?  Suppose every byte of LONGWORD has a bit set
    1.70 +-	 somewhere.  There will be a carry into bit 8.  If bit 8
    1.71 +-	 is set, this will carry into bit 16.  If bit 8 is clear,
    1.72 +-	 one of bits 9-15 must be set, so there will be a carry
    1.73 +-	 into bit 16.  Similarly, there will be a carry into bit
    1.74 +-	 24.  If one of bits 24-30 is set, there will be a carry
    1.75 +-	 into bit 31, so all of the hole bits will be changed.
    1.76 +-
    1.77 +-	 The one misfire occurs when bits 24-30 are clear and bit
    1.78 +-	 31 is set; in this case, the hole at bit 31 is not
    1.79 +-	 changed.  If we had access to the processor carry flag,
    1.80 +-	 we could close this loophole by putting the fourth hole
    1.81 +-	 at bit 32!
    1.82 +-
    1.83 +-	 So it ignores everything except 128's, when they're aligned
    1.84 +-	 properly.  */
    1.85 +-
    1.86 +       longword = *longword_ptr++;
    1.87 + 
    1.88 +-      if (
    1.89 +-#if 0
    1.90 +-	  /* Add MAGIC_BITS to LONGWORD.  */
    1.91 +-	  (((longword + magic_bits)
    1.92 +-
    1.93 +-	    /* Set those bits that were unchanged by the addition.  */
    1.94 +-	    ^ ~longword)
    1.95 +-
    1.96 +-	   /* Look at only the hole bits.  If any of the hole bits
    1.97 +-	      are unchanged, most likely one of the bytes was a
    1.98 +-	      zero.  */
    1.99 +-	   & ~magic_bits)
   1.100 +-#else
   1.101 +-	  ((longword - lomagic) & himagic)
   1.102 +-#endif
   1.103 +-	  != 0)
   1.104 ++      /* This hack taken from Alan Mycroft's HAKMEMC postings.
   1.105 ++         See: http://www.cl.cam.ac.uk/~am21/progtricks.html */
   1.106 ++      if (((longword - lomagic) & ~longword & himagic) != 0)
   1.107 + 	{
   1.108 +-	  /* Which of the bytes was the zero?  If none of them were, it was
   1.109 +-	     a misfire; continue the search.  */
   1.110 + 
   1.111 + 	  const char *cp = (const char *) (longword_ptr - 1);
   1.112 +