Facundo Ciccioli writes:
> FaQ
>
> 2007/3/29, Andrew Haley <aph@xxxxxxxxxx>:
> > Facundo Ciccioli writes:
> > >
> > > This is obviusly not urgent, since the first code is perfectly
> > > acceptable and applicable to what I am doing, but I just got curious.
> >
> > This is an arithmetic overflow, and is explicitly undefined. One
> > result is as good as any other. What do you expect the code to do?
> >
> I expect to get a 1 in the s-th bit of a, and zeroes in all other
> bits. I don't know why you say that's an arithmetic overflow, if
> unsigned long long is 64 bits long and it has a 63-th bit (the last
> one).
>
> Anyway, Rupert Wood comment works, thanks a lot.
So, even after Rupert Wood's explanation, you still don't realize
that you had written an arithmetic overflow? Let's have a look at
your code:
int main() {
unsigned long long a;
unsigned s= 63;
a= (1 << s);
return 0;
}
This expression is a 32-bit signed integer constant 1 shifted left 63
places, which overflows:
(1 << s);
You then assign the result of this overflow to the 64-bit signed
integer variable a:
a= (1 << s);
Andrew.
--
Red Hat UK Ltd, Amberley Place, 107-111 Peascod Street, Windsor, Berkshire, SL4 1TE, UK
Registered in England and Wales No. 3798903
