[ILUG] integer overflow exception

[Jeroen Massar]

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Pádraig Brady wrote:
> I was wondering under what circumstances
> (compilers/CPUs) that the following program
> generates an integer overflow exception.
> I think none, but I'm reviewing code
> that assumes otherwise.
> I understand MIPS generates an exception
> for signed arithmetic only.
> 
> #include <stdio.h>
> int main(void)
> {
>     unsigned int i;
>     for(;;i++;) {
>         if (i==0) {
>             printf("wrap\n");
>             fflush(stdout);
>         }
>     }
> }

IMHO that could doesn't work anyway. It is not mandatory for a
compiler to initialize any variables, as such the above code will be
random. Though it does accomplish your mission.

The only way to avoid this is to always test after or before an
increase if your value still fits inside the maximums. That works on
all platforms and on all cases. Easiest way is:

unsigned int prev_i = i;
i++;
if (i < prev_i) printf("wrapped!");

I actually am pretty sure a lot of software does not account for this.
Lots of people (me included :) simply use uint64_t ;)
They can wrap, but it should not quickly happen.
If you really want to avoid that you can always use a long double or
something which will nicely scale away the larger it gets.

All depends for what application you need this.

Greets,
 Jeroen
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Comment: Jeroen Massar / http://unfix.org/~jeroen/

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