Leonitis wrote:
> Thanks for replying!
>
> However, what I want to happen is for the product to still be a 32-bit
> integer.
>
> My problem is that GCC takes uses the upper 16 bits of 'a' and 'b' as well
> as the lower bits. If you multiply the upper 16 bits of 'a' with the lower
> 16 bits of 'b', then you still get a number that is contained within
> 32-bits, and hence it affects the product that is added to the accumulate
> value.
Right, so you *must* prevent the overflow. Why do you need the signed
overflow anyway? Can't you use unsigned arithmetic?
Like this:
unsigned int c;
unsigned short a = 0;
unsigned short b = 0;
int count;
for(count = 0; count < cycles; count++, a++, b++)
{
c += a * b ;
}
Andrew.
