What x = n++; would do is assign the value of n immediately to x, and then increment n by 1.
What x = ++n; would do is increment n by 1 immediately, and then assign that new value to x.

I'm still not grasping this fully. From what I gather the resulting x is going to be the same value either way, correct ? So then why would you need the two different ways (++n and n++) to achieving it ?

Can anyone explain this for me ?

Thanks,
Tony

Your understanding is correct, but your conclusion is not. Perhaps it is easier if you think of these in real-life terms. Let's make 'n' a variable, with a value of "10".

So...
x = $n++ (post-increment operator)
This makes x=10 and $n=11

x = ++$n (pre-increment operator)
This makes x=11 and $n=11

Make sense?

Thanks CS2,

I understand how it's done now.

Tony