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

[Daniel Fischer]

Am Freitag, 27. Februar 2009 22:00 schrieb Anonymous Anonymous:
> 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:

Not quite. The type of until, written with one parenthesis which is not 
strictly necessary, is

until :: (a -> Bool) -> (a -> a) -> (a -> a)

Now unify that with the type of the first argument of

foldr :: (s -> t -> t) -> t -> [s] -> t

You get

s = a -> Bool
t = a -> a

and

foldr until :: t -> [s] -> t, which expands to

foldr until :: (a -> a) -> [a -> Bool] -> (a -> a)

And if you don't want to do the analysis yourself,
Prelude> :t foldr until
foldr until :: (a -> a) -> [a -> Bool] -> a -> a


>
> 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".

I can't really see a useful application of foldr until, sorry.
But
Prelude> foldr until (+1) [(> 10), (> 12)] 4
13

However:
Prelude> foldr until (+1) [(> 10), (>8)] 4
Interrupted.


>
> Thanks!