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>> 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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