1 Original patch from: gentoo/src/patchsets/glibc/2.9/1020_all_glibc-2.9-strlen-hack.patch
3 -= BEGIN original header =-
4 http://sourceware.org/bugzilla/show_bug.cgi?id=5807
5 http://www.cl.cam.ac.uk/~am21/progtricks.html
7 -= END original header =-
9 diff -durN glibc-2_9.orig/string/strlen.c glibc-2_9/string/strlen.c
10 --- glibc-2_9.orig/string/strlen.c 2005-12-14 12:09:07.000000000 +0100
11 +++ glibc-2_9/string/strlen.c 2009-02-02 22:00:51.000000000 +0100
15 const unsigned long int *longword_ptr;
16 - unsigned long int longword, magic_bits, himagic, lomagic;
17 + unsigned long int longword, himagic, lomagic;
19 /* Handle the first few characters by reading one character at a time.
20 Do this until CHAR_PTR is aligned on a longword boundary. */
22 if (*char_ptr == '\0')
23 return char_ptr - str;
25 - /* All these elucidatory comments refer to 4-byte longwords,
26 - but the theory applies equally well to 8-byte longwords. */
28 longword_ptr = (unsigned long int *) char_ptr;
30 - /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
31 - the "holes." Note that there is a hole just to the left of
32 - each byte, with an extra at the end:
34 - bits: 01111110 11111110 11111110 11111111
35 - bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
37 - The 1-bits make sure that carries propagate to the next 0-bit.
38 - The 0-bits provide holes for carries to fall into. */
39 - magic_bits = 0x7efefeffL;
40 himagic = 0x80808080L;
41 lomagic = 0x01010101L;
42 if (sizeof (longword) > 4)
44 /* 64-bit version of the magic. */
45 /* Do the shift in two steps to avoid a warning if long has 32 bits. */
46 - magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
47 himagic = ((himagic << 16) << 16) | himagic;
48 lomagic = ((lomagic << 16) << 16) | lomagic;
51 if *any of the four* bytes in the longword in question are zero. */
54 - /* We tentatively exit the loop if adding MAGIC_BITS to
55 - LONGWORD fails to change any of the hole bits of LONGWORD.
57 - 1) Is this safe? Will it catch all the zero bytes?
58 - Suppose there is a byte with all zeros. Any carry bits
59 - propagating from its left will fall into the hole at its
60 - least significant bit and stop. Since there will be no
61 - carry from its most significant bit, the LSB of the
62 - byte to the left will be unchanged, and the zero will be
65 - 2) Is this worthwhile? Will it ignore everything except
66 - zero bytes? Suppose every byte of LONGWORD has a bit set
67 - somewhere. There will be a carry into bit 8. If bit 8
68 - is set, this will carry into bit 16. If bit 8 is clear,
69 - one of bits 9-15 must be set, so there will be a carry
70 - into bit 16. Similarly, there will be a carry into bit
71 - 24. If one of bits 24-30 is set, there will be a carry
72 - into bit 31, so all of the hole bits will be changed.
74 - The one misfire occurs when bits 24-30 are clear and bit
75 - 31 is set; in this case, the hole at bit 31 is not
76 - changed. If we had access to the processor carry flag,
77 - we could close this loophole by putting the fourth hole
80 - So it ignores everything except 128's, when they're aligned
83 longword = *longword_ptr++;
87 - /* Add MAGIC_BITS to LONGWORD. */
88 - (((longword + magic_bits)
90 - /* Set those bits that were unchanged by the addition. */
93 - /* Look at only the hole bits. If any of the hole bits
94 - are unchanged, most likely one of the bytes was a
98 - ((longword - lomagic) & himagic)
101 + /* This hack taken from Alan Mycroft's HAKMEMC postings.
102 + See: http://www.cl.cam.ac.uk/~am21/progtricks.html */
103 + if (((longword - lomagic) & ~longword & himagic) != 0)
105 - /* Which of the bytes was the zero? If none of them were, it was
106 - a misfire; continue the search. */
108 const char *cp = (const char *) (longword_ptr - 1);