Although I doubt anyone would write it, this would work, wouldn't it?
int n = +10;
while ( n > 0)
n += -1;
Bill
> On 16 February 2011 20:09, Jason Mancini wrote:
> >
> > Hello,
> > So as I recall, the following can be an infinite loop now with optimizations, right?
> >
> > for (int i(1); i!=0; ++i) { ... }
>
> Right.
>
> > What about:
> >
> > unsigned int x = 0xFFFFFFFFU;
> > x = x+1;
> > if (x) { ... can we get here because "positive x + 1 must still positive"? ... }
> >
> > If not, given the first, why not?
>
> No. The C and C++ standards define that unsigned integers do not
> overflow, they wrap, with well-defined behaviour.
>
> They do not define what happens if a signed integer overflows, so your
> first loop results in undefined behaviour, and so you cannot
> reasonably expect any particular behaviour. The compiler can do
> whatever it likes with your code.
>
> Put another way:
> There is no way for a correct C or C++ program to increment a signed
> integer greater than zero such that the result is zero. Because a
> correct C or C++ program does not contain integer overflows.
>
>
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