On Tue, Dec 30, 2008 at 05:32:57PM +0100, grischka wrote:
> Christian Jullien wrote:
>> printf("fails %08x\n", (~0) >> 1);
> fails ffffffff
> Maybe gcc is not correct but then I need an exact explanation why ;)
0 has type int. On x86 the representation is 32 value bits set to
zero, with no padding bits.
~0 flips all of the bits of the representation, producing an int with
value -1.
(~0) >> 1 is a right-shift of a signed integer with negative value,
which is an implementation-defined operation. On x86, this is usually
implemented with the SAR instruction, which divides by 2 with rounding
toward -infinity. -1/2 rounded toward -infinity is -1, so the value
is unchanged.
This then passes int -1 to a printf specifier that expects unsigned
int. Since the value is negative and cannot be represented by
unsigned int, you get undefined behavior and any result is allowed.
I believe in this case, the representation of int -1 is reinterpreted
as an unsigned int, producing the printed value ffffffff. But really,
the test itself is fundamentally broken.
-Dave Dodge
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