On Mon, Jun 2, 2008 at 9:48 PM, Scott Phuong <mycleanjunk@xxxxxxxxx> wrote:
>
> unsigned short a;
> unsigned short b;
>
> a = 0xFFFF;
> b = 0x3FC;
>
> a = (a + 1) % b;
> printf ("A is 0x%x\n", a);
> // I expect the answer to be 0 and it is not! It is 0x100. Why is this?
>
I bet if you did
++a; a %= b;
you'd get 1.
I don't know the exact rules, but 1 is an int, so a+1 will give you
(int)a + 1, which will be 0x10000, which when modded by (int)b will
not be 0.
I'd have to read up on integral promotion to be sure, but I think the
only way is to add an explicit cast.
a = (unsigned short)(a+1) % b;
(If you want to keep it one expression, and of that form. ++a, a %=b;
is possible, as mentioned, and a = ( (a+1) & 0xFFFFu ) % b would also
work.)
HTH,
~ Scott
