7 Dec
2000
7 Dec
'00
12:23 p.m.
| < p y = (let f :: c -> Bool; f x = op (y >> return x) in f, y ++ []) | < q y = (y ++ [], let f :: c -> Bool; f x = op (y >> return x) in f) | < | System CT accepts both. No type signature is required. | | The types inferred for p and q are, resp.: | | Forall a,b. [a] -> (b -> Bool, [a]) and | Forall a,b. [a] -> ([a], b -> Bool) That's not the type we wanted, though. We wanted Forall a. [a] -> (Forall b. b->Bool, [a]) and similarly the other one. That's the reason for the type signature on f, which is short for (forall c. c->Bool). If you leave out the type signature then all Haskell compilers are happy. (Of course, this isn't a very useful example, but it's the smallest one that shows up the problem.) Simon