[Haskell-cafe] How can you effectively manually determine the
type?
Dan Weston
westondan at imageworks.com
Fri Feb 27 16:05:37 EST 2009
Does writing it like this help any?
until :: (c -> Bool) -> (c -> c) -> (c -> c)
foldr :: (( a ) -> ( b ) -> ( b )) -> b -> [a] -> b
Anonymous Anonymous wrote:
> Hello,
>
> I'm new at haskell and I have the following question:
>
> let's say I type the following:
>
> function = foldr until
>
> Now my first question is what is the type of this function? Well let's
> see what the type of until and foldr is:
>
> until :: (a -> Bool) -> (a -> a) -> a -> a
> foldr :: (a -> b -> b) -> b -> [a] -> b
>
> So I would be thinking: we fill until in the position of (a -> b -> b)
> so, a correspond with (a -> Bool) and b correspond with (a -> a) and b
> correspond with a. Hmm a small problem, I think we can divide that as
> follows: b1 corresponds with (a -> a) and b2 corresponds with a. So I get:
>
> foldr until :: b1 -> [a] -> b2
> foldr until :: (a -> a) -> [a -> Bool] -> a
>
> Is this a correct way of thinking or am I wrong?
>
> And another question is: can someone give me an example how this can be
> executed? All my code that I tried to execute resulted in errors with
> "foldr until".
>
> Thanks!
>
>
More information about the Haskell-Cafe
mailing list