# [Haskell-cafe] Re: Function Type Calculation (Take 2)

PR Stanley prstanley at ntlworld.com
Fri May 2 03:43:34 EDT 2008

```Just in case anyone missed this:
[1]
funk f x = f (funk f) x

f :: a
x :: b
funk f x :: c
therefore funk :: a -> b -> c

RHS
f (funk f) x :: c

f (funk f) :: d -> c
x :: d

f :: e -> d -> c

funk :: h -> e
f :: h

unification
f :: a = h = (e -> d -> c)
x b = d

No. x :: b = d (a typo?)
Paul: What's wrong with x being of type b and of type d? Could you

Don't forget also that

funk :: a -> b -> c = h -> e,

which means that e = b -> c
Paul: is that something to do with partial application? (funk f) is
a partially applied function, correct? Again an explanation would be
appreciated.

therefore funk :: ((h -> e) -> b -> c) -> b -> c

No. I don't understand where you've got this expression from. It's

funk :: a -> b -> c = (e -> d -> c) -> b -> c = ((b -> c) -> b -> c) - > b -> c

According to GHCi:

Prelude> let funk f x = f (funk f) x
Prelude> :t funk
funk :: ((t1 -> t) -> t1 -> t) -> t1 -> t

Thanks
Paul

```