crosstool-ng/patches/glibc/2.9/210-2.9-strlen-hack.patch

110 lines
4.0 KiB
Diff

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);