Re: Funny loop with post-increment.

Silky:
> it's not "pre" i++. it happens after that. i is equal to the result of
> the expression "i++". what you are missing is that i++ in that
> expression doesn't actually affect the end result of "i" (even though
> it is calculated) because it's overwritten.

I'm not missing anything any more, thank you everyone, but I figured it out
long before I could properly describe it, except for this reply to Jon:

> i is 0
>
> <begin i++>
> tmp = i
> i += 1
> <end i++>
>
> i = tmp

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