infinite for-loop and related question

[Bill McEnaney]

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?
> >
> > Â  for (int i(1); i!=0; ++i) { ... }
> 
> Right.
> 
> > What about:
> >
> > Â  unsigned int x = 0xFFFFFFFFU;
> > Â  x = x+1;
> > Â  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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