Sorry, am subscribed with digest and haven't got David House's reply via email yet, hence the broken thread. It's not to do with lambdas though:
:t let f r s = let g (fn::forall n . (Num n) => n -> n) = return (fn r, fn s) in (return negate) >>= g in f
<interactive>:1:98: Couldn't match expected type `a -> a' against inferred type `forall n. (Num n) => n -> n' Expected type: (a -> a) -> m b Inferred type: (forall n2. (Num n2) => n2 -> n2) -> m1 (n, n1) In the second argument of `(>>=)', namely `g' In the expression: let g (fn :: forall n. (Num n) => n -> n) = return (fn r, fn s) in (return negate) >>= g ...unless we're talking about lambdas in the definitions of return and
= which would be quite worrying.
Matthew -- Matthew Sackman http://www.wellquite.org/