Hi,
For the following program
> foo :: (Int, String, t -> t) -> (Int, String)
> foo (i, s, f) = (f i, f s)
ghc 0.24 reports
"foo.hs", line 2:
Couldn't match the type `Int' against `String'.
In the first argument of `f', namely `s',
Expected type of the argument: Int
Inferred type of the argument: String
Possible cause of error: `f' is not polymorphic
it is monomorphic in the type variable(s): o_4
(nice error message, BTW).
It would be nice if you could explicitly declare that `f'
had a polymorphic higher-order type `forall t . t -> t',
e.g. by
> foo :: (Int, String, (forall t . t -> t)) -> (Int, String)
However, the Haskell report says that there is no way to declare an
explicitly universally quantified type. My question is simple: why not?
--
Fergus Henderson <[EMAIL PROTECTED]> | "I have always known that the pursuit
WWW: <http://www.cs.mu.oz.au/~fjh> | of excellence is a lethal habit"
PGP: finger [EMAIL PROTECTED] | -- the last words of T. S. Garp.
