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 [EMAIL PROTECTED] 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
Virus checked by G DATA AntiVirusKit
Version: AVK 12.0.37 from 06.12.2002
Virus news: www.antiviruslab.com
___
Haskell mailing list
Haskell@haskell.org
http://www.haskell.org/mailman/listinfo/haskell
___
Haskell mailing list
Haskell@haskell.org
http://www.haskell.org/mailman/listinfo/haskell
