code questions.

code questions.

[Scott Phuong]

Hi GCC Help,

What is my confusion here? I am using gcc 4.1.2-14 on an i386 machine.
I don't know why I am getting a different value than I expect. Is
there a compiler flag I can use to make the expression evaluate
correctly according to my expectations? Thanks!

int main(char argc, char* agrv[]) {

	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?

	a = 0xFFFF;

	a = (a + 1);
	printf ("A is 0x%x\n", a);
	// I expect the answer to be 0 and it is.

	a = a % b;
	printf ("A is 0x%x\n", a);
	// I expect the answer to be 0 and it is.

}

Best Regards,

Scott

Re: code questions.

[me22]

On Mon, Jun 2, 2008 at 9:48 PM, Scott Phuong <mycleanjunk@gmail.com> 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

Re: code questions.

[Scott Phuong]

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@gmail.com> wrote:
> On Mon, Jun 2, 2008 at 9:48 PM, Scott Phuong <mycleanjunk@gmail.com> 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
>

Re: code questions.

[me22]

On Mon, Jun 2, 2008 at 10:04 PM, Scott Phuong <mycleanjunk@gmail.com> wrote:
> 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?

Switch to a platform with 16-bit ints?

> 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.
>

More evidence that you're hitting integral promotion, so there's
basically nothing you can do about it.

(Besides make sure a is never that large :P)

Re: code questions.

[Scott Phuong]

I see. Well, thank you very much for your help. I appreciate the quick response.

Scott

On Mon, Jun 2, 2008 at 7:11 PM, me22 <me22.ca@gmail.com> wrote:
> On Mon, Jun 2, 2008 at 10:04 PM, Scott Phuong <mycleanjunk@gmail.com> wrote:
>> 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?
>
> Switch to a platform with 16-bit ints?
>
>> 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.
>>
>
> More evidence that you're hitting integral promotion, so there's
> basically nothing you can do about it.
>
> (Besides make sure a is never that large :P)
>

Re: code questions.

[Tony Wetmore]

Scott,

I just tried the following:

   a = (unsigned short) ( a + 1 ) % b;

which appears to work as you expect.  I tried it using gcc 3.4.4 on 
Windows (Cygwin).

Good luck!

--
Tony Wetmore
Raytheon Solipsys


Scott Phuong wrote:
> I see. Well, thank you very much for your help. I appreciate the quick response.
> 
> Scott
> 
> On Mon, Jun 2, 2008 at 7:11 PM, me22 <me22.ca@gmail.com> wrote:
>> On Mon, Jun 2, 2008 at 10:04 PM, Scott Phuong <mycleanjunk@gmail.com> wrote:
>>> 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?
>> Switch to a platform with 16-bit ints?
>>
>>> 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.
>>>
>> More evidence that you're hitting integral promotion, so there's
>> basically nothing you can do about it.
>>
>> (Besides make sure a is never that large :P)
>>
> 
>