Re: Implementation-defined behaviour

[Philip Potter]

Ioannis Vranos wrote:
> So, printf("%d "%d\n", x++, x++); invokes implementation-defined
> behaviour, while printf("%d\n", x++); doesn't invoke
> implementation-defined behaviour?
Aside from the fact that this won't compile (you have one too many " 
characters), this is almost exactly the same question as FAQ 3.2. The 
problem is not the individual 'x++' expressions, but the way they 
interact with each other.

Philip