On 5/27/07, Andrew Coppin <[EMAIL PROTECTED]> wrote:
map :: (a -> b) -> [a] -> [b]
map :: (a -> b) -> ([a] -> [b])
Which is beautifully symmetric. Alternatively, you can think about how
you actually use it:
map :: ((a -> b) -> [a]) -> [b]
No, now you're confusing things. The uncurried function looks like this:
uncurry map :: (a -> b, [a]) -> [b]
Note that map :: ( .... ) -> [b] only takes one argument, not two
HTH
/Thomas
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