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[BZ #5807]

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2009-03-15  Ulrich Drepper  <drepper@redhat.com>
	[BZ #5807]
	* string/strlen.c (strlen): Fix omission in the expression to test
	for NUL bytes.
This commit is contained in:
Ulrich Drepper
2009-03-15 09:03:38 +00:00
parent fad070abc4
commit 71a5bd3e17
4 changed files with 36 additions and 151 deletions
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52
string/strlen.c
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@ -1,4 +1,4 @@
/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc.
/* Copyright (C) 1991,1993,1997,2000,2003,2009 Free Software Foundation, Inc.
This file is part of the GNU C Library.
Written by Torbjorn Granlund (tege@sics.se),
with help from Dan Sahlin (dan@sics.se);
@ -32,7 +32,7 @@ strlen (str)
{
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. */
@ -56,14 +56,12 @@ strlen (str)
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;
}
@ -75,53 +73,9 @@ strlen (str)
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)
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. */