G'day all.
Ketil Malde wrote:
> (.) . (.) .(.)
>
> I entered it into GHCi, and got
>
> :: forall a a b c a.
> (b -> c) -> (a -> a -> a -> b) -> a -> a -> a -> c
I got this:
Prelude> :t (.) . (.) . (.)
(.) . (.) . (.) :: forall a a1 b c a2.
(b -> c) -> (a -> a1 -> a2 -> b) -> a -> a1 -> a2 -> c
Quoting Tim Docker <[EMAIL PROTECTED]>:
> Is there a straightforward technique by which type signatures can be
> "hand calculated", or does one end up needing to run the whole inference
> algorithm in one's head?
For this case, yes, but you need to expand the point-free style first.
Using the combinator B to represent (.):
(.) . (.) . (.)
= \f -> B (B (B f))
= \f g x -> B (B (B f)) g x
= \f g x -> B (B f) (g x)
= \f g x y -> B (B f) (g x) y
= \f g x y -> B f (g x y)
= \f g x y z -> f (g x y z)
The type should now be obvious.
However, it's probably easier in this case to get the expanded lambda
expression directly from the type's "free theorem".
Cheers,
Andrew Bromage
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