On 2003-11-18 at 10:46EST "Abraham Egnor" wrote: > The classic way to write a lift function for tuples is, of course: > > liftTup f (a, b) = (f a, f b) > > which has a type of (a -> b) -> (a, a) -> (b, b). I've been wondering if > it would be possible to write a function that doesn't require the types in > the tuple to be the same, just that the types in the second tuple are the > result of applying the type transformation implied in the function to be > lifted to the types in the first tuple. Now, in Haskell98, this isn't > possible because of the monomorphism restriction; however, ghc > conveniently has a way to disable that. However, I'm still having > problems figuring out if it's even doable within the current constraints > of the glasgow-extended type system. One possibility I tried is: > > liftTup (f :: forall a b. a -> b) (p, q) = (f p, f q)
Note that the type you are requiring for f is equivalent to forall a . a -> forall b . b which rather limits the possible values for f. Another possibility would be lifTup:: (forall a. a -> b) -> (c, d) -> (b,b) but that requires the f to be polymorphic and that the result has elements of the same type. What you want is that f be applicable to both b and c, giving results b' and c', but if b and c happen to be the same type then f need not be polymorphic. I don't think you can express this in ghc's type system. You'd have to have bounded quantification: lifTup :: forall c, a >= c, b >= c, d, a'<=d, b'<= d. (c -> d) -> (a,b)->(a',b') which is monstrous even if I've managed to get it right! Jón -- Jón Fairbairn [EMAIL PROTECTED] _______________________________________________ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
