On 08/05/07, Matthew Sackman
:t let f r s = (return negate) >>= (\(fn::forall n . (Num n) => n -> n) -> return (fn r, fn s)) in f
<interactive>:1:35: Couldn't match expected type `a -> a' against inferred type `forall n. (Num n) => n -> n' In the pattern: fn :: forall n. (Num n) => n -> n In a lambda abstraction: \ (fn :: forall n. (Num n) => n -> n) -> return (fn r, fn s) In the second argument of `(>>=)', namely `(\ (fn :: forall n. (Num n) => n -> n) -> return (fn r, fn s))'
I.e. why does the polymorphism get destroyed?
Here fn is bound by a lambda abstraction, and is therefore monomorphic. I can't find anything in the Report about that, but that is how it works. It might be how a H-M type system works in general, I'm not sure. -- -David House, dmhouse@gmail.com