>> I bet you can't overload ";", can you?
> 
> ';' is not an operator. It is a statement terminator.

Yes, I know that. ;-) I was just being silly...

>> In Haskell, you can. >:-D
> 
> Can you overload/redefine '->' in Haskell?

No. But then, it means something different in Haskell anyway.

> What about '()'?

No.

But then, in Haskell, a function *is* a data value, so there's not so 
much need...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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