On Sun, Nov 16, 2008 at 1:32 PM, Maurício <[EMAIL PROTECTED]> wrote: > Hi, > > Why is this wrong? > > ---- > class MyClass r where function :: r -> s > > data MyData u = MyData u > > instance MyClass (MyData v) where function (MyData a) = a > ---- > > GHC says that the type of the result of 'function' is both determined by > the "rigid type" from MyClass and the "rigid type" from MyData. But why > can't both be the same?
As Bulat said, your type signature is equivalent to: function :: forall r s. r -> s whereas the result you're producing can't produce any s, but only particular s's. In essence, the result type is determined by the input type. One way to code this would be to use functional dependencies: class MyClass r s | r -> s where function :: r -> s data MyData u = MyData u instance MyClass (MyData v) v where function (MyData a) = a /g -- I am in here _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
