[Haskell-cafe] How can you effectively manually determine the type?

[Dan Weston]

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!
>  
>