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>>> I bet you can't overload ";", can you? In Haskell, you can. >:-D
>>
>> ; is not an operator in the first place.
>
> Yeah, I know that. ;-) I'm just teasing.
Wow, wait a sec...
http://www.research.att.com/~bs/whitespace98.pdf
WTF?! O_O
--
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