Hi,
Thanks so much for your quick reply. This code is simliar to code in a
third party source code that I have to fix and they do this all over
the place. The question is w/o touching their code is there a way I
can fix this? I tried explicitly casting the 1 to an unsigned short
but it doesn't seem to work like this:
a = (a + (unsigned short) 1) % b.
Thanks for your help.
Scott
On Mon, Jun 2, 2008 at 6:59 PM, me22 <me22.ca@xxxxxxxxx> wrote:
> 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
>
