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

[PR Stanley]

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 
perhaps explain the error please?

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

which is about the same.


Thanks
Paul