Using "type"

[Simon Peyton-Jones]

GHC currently takes the following (non-Haskell 98) attitude to type
synonyms:

	first expand type synonyms
	only then check for partial applications of type synonyms
	then perform type checking

So in his case type for f expands to:

	f :: m () -> m ()

	test :: Int -> Int

You can see how the type for test comes out:
	Thing (Const Int) =3D Const Int () =3D Int

Now GHC tries to unify=20
	m ()   with    Int

and fails, as it should.   The error message is phrased in terms
of the original un-expanded types, which is usually a good thing but
on this occasion gives a rather odd looking error message.

GHC shouldn't really do this lazy checking unless you say -fglasgow-exts

Simon

| -----Original Message-----
| From: Samuel E. Moelius III [mailto:usmoeliu@mcs.drexel.edu]=20
| Sent: 04 May 2002 20:45
| To: haskell-cafe@haskell.org
| Subject: Using "type"
|=20
|=20
| I apologize if this question has been asked before...
|=20
| It seems to me the following code should be legal:
|=20
| 	type Thing m =3D m ()
| =09
| 	type Const a b =3D a
| =09
| 	f :: Thing m -> Thing m
| 	f x =3D x
| =09
| 	test :: Thing (Const Int) -> Thing (Const Int)
| 	test =3D f
|=20
| However, GHC gives the following error:
|=20
|      Couldn't match `Thing m' against `Thing (Const Int)'
|          Expected type: Thing m
|          Inferred type: Thing (Const Int)
|          Expected type: Thing (Const Int) -> Thing (Const Int)
|          Inferred type: Thing m -> Thing m
|=20
| But if you change the definition of Const
|=20
| 	newtype Const a b =3D MakeConst a
|=20
| the code compiles fine.
|=20
| Is this the desired behavior?
|=20
| Is it the case that when you have quantification of a type=20
| variable of=20
| kind * -> * (in this case, m), that variable may only become=20
| bound to a=20
| type constructor?
|=20
| Sam Moelius
|=20
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|=20