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!
_______________________________________________
Haskell-Cafe mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/haskell-cafe
