On Sat 19/04/08 9:57 PM , Christian Böhme monodhs@xxxxxx sent:
> Hi all,
>
>
>
> I ran into a problem with an expression that I imagined is too
>
> trivial to even needing a second thought in the first place.
>
> However, on a 32 bit machine, the code
>
>
>
> --->8---
>
>
>
> #include
>
>
>
> // ...
>
>
>
> int64_t a;
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> uint32_t b = 8;
>
>
>
> a = -(b * 2u);
>
>
>
> --->8---
>
>
>
> assigns a the value 0xfffffff0 while no warnings were issued even
>
> when compiling with all warnings (-Wall) switched on. Strangely
>
> enough and for reasons escaping me
What type is the expression -(b * 2u)? And why would the number 2^32 - 16 be
sign extended when stored in a 64-bit signed int? 2^32 - 16 is a valid positive
number for the 64-bit type.
Consider
int main(void)
{
long long a;
unsigned long b;
b = 8;
a = (int)(-(b * 2U));
printf("a == %lld\n", a);
return 0;
}
Why does this work when your example does not?
Tom
