[Haskell] type question revisited

[Abraham Egnor]

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 at 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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