Your first attempt didn't typecheck simply because

in return () means that the return value of the function is monadic, but you did not declare as such.

In your second version, the type of shf is *not* a rank-2 type; it's exactly the same type as shw. This can be expressed (with ghc extensions) as

foo :: (a -> String) -> [a] -> [String] foo (shw :: t) x = let shf :: t shf o = shw o in map shf x

or equivalently

foo :: (a -> String) -> [a] -> [String] foo (shw :: t -> String) x = let shf :: t -> String shf o = shw o in map shf x

The essential aspect is that the 'a' from the type signature is *not* in scope for the let-bound type signature; you have to bring the appropriate variable 't' in by using an in-line type signature for 'shw'. Abe On 6/3/05, mv <mv42c@dodo.com.au> wrote:

I answered my own question only to raise another - what I wanted to do is this

...

foo :: (a -> String) -> [a] -> [String] foo shw x = let shf :: ( forall a . a ) -> String shf o = shw o

in map shf x

the type of shf is a rank 2 type - but how do you map it ? as the above gives thise error in hugs:

Use of shf requires at least 1 argument

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