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110 lines
4.0 KiB
Diff
110 lines
4.0 KiB
Diff
Original patch from: gentoo/src/patchsets/glibc/2.9/1020_all_glibc-2.9-strlen-hack.patch
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-= BEGIN original header =-
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http://sourceware.org/bugzilla/show_bug.cgi?id=5807
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http://www.cl.cam.ac.uk/~am21/progtricks.html
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-= END original header =-
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diff -durN glibc-2_9.orig/string/strlen.c glibc-2_9/string/strlen.c
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--- glibc-2_9.orig/string/strlen.c 2005-12-14 12:09:07.000000000 +0100
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+++ glibc-2_9/string/strlen.c 2009-02-02 22:00:51.000000000 +0100
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@@ -32,7 +32,7 @@
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{
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const char *char_ptr;
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const unsigned long int *longword_ptr;
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- unsigned long int longword, magic_bits, himagic, lomagic;
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+ unsigned long int longword, himagic, lomagic;
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/* Handle the first few characters by reading one character at a time.
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Do this until CHAR_PTR is aligned on a longword boundary. */
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@@ -42,28 +42,14 @@
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if (*char_ptr == '\0')
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return char_ptr - str;
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- /* All these elucidatory comments refer to 4-byte longwords,
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- but the theory applies equally well to 8-byte longwords. */
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-
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longword_ptr = (unsigned long int *) char_ptr;
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- /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits
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- the "holes." Note that there is a hole just to the left of
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- each byte, with an extra at the end:
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-
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- bits: 01111110 11111110 11111110 11111111
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- bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD
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-
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- The 1-bits make sure that carries propagate to the next 0-bit.
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- The 0-bits provide holes for carries to fall into. */
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- magic_bits = 0x7efefeffL;
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himagic = 0x80808080L;
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lomagic = 0x01010101L;
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if (sizeof (longword) > 4)
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{
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/* 64-bit version of the magic. */
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/* Do the shift in two steps to avoid a warning if long has 32 bits. */
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- magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;
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himagic = ((himagic << 16) << 16) | himagic;
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lomagic = ((lomagic << 16) << 16) | lomagic;
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}
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@@ -75,56 +61,12 @@
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if *any of the four* bytes in the longword in question are zero. */
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for (;;)
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{
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- /* We tentatively exit the loop if adding MAGIC_BITS to
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- LONGWORD fails to change any of the hole bits of LONGWORD.
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-
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- 1) Is this safe? Will it catch all the zero bytes?
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- Suppose there is a byte with all zeros. Any carry bits
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- propagating from its left will fall into the hole at its
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- least significant bit and stop. Since there will be no
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- carry from its most significant bit, the LSB of the
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- byte to the left will be unchanged, and the zero will be
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- detected.
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-
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- 2) Is this worthwhile? Will it ignore everything except
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- zero bytes? Suppose every byte of LONGWORD has a bit set
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- somewhere. There will be a carry into bit 8. If bit 8
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- is set, this will carry into bit 16. If bit 8 is clear,
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- one of bits 9-15 must be set, so there will be a carry
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- into bit 16. Similarly, there will be a carry into bit
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- 24. If one of bits 24-30 is set, there will be a carry
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- into bit 31, so all of the hole bits will be changed.
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-
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- The one misfire occurs when bits 24-30 are clear and bit
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- 31 is set; in this case, the hole at bit 31 is not
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- changed. If we had access to the processor carry flag,
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- we could close this loophole by putting the fourth hole
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- at bit 32!
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-
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- So it ignores everything except 128's, when they're aligned
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- properly. */
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-
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longword = *longword_ptr++;
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- if (
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-#if 0
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- /* Add MAGIC_BITS to LONGWORD. */
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- (((longword + magic_bits)
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-
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- /* Set those bits that were unchanged by the addition. */
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- ^ ~longword)
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-
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- /* Look at only the hole bits. If any of the hole bits
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- are unchanged, most likely one of the bytes was a
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- zero. */
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- & ~magic_bits)
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-#else
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- ((longword - lomagic) & himagic)
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-#endif
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- != 0)
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+ /* This hack taken from Alan Mycroft's HAKMEMC postings.
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+ See: http://www.cl.cam.ac.uk/~am21/progtricks.html */
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+ if (((longword - lomagic) & ~longword & himagic) != 0)
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{
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- /* Which of the bytes was the zero? If none of them were, it was
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- a misfire; continue the search. */
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const char *cp = (const char *) (longword_ptr - 1);
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