On 19 July 2015 at 10:13, Clinton Mead wrote:

Lets say I've got the following data type:

data D c a b = D (c a b) (c b a)

And I define a function to manipulate it:

apply :: (c a b -> c a2 b2) -> (c b a -> c b2 a2) -> D c a b -> D c a2 b2 apply f1 f2 (D x y) = D (f1 x) (f2 y)

This is all fine. But I want a shorter function if (f1 = f2). So I write:

applyBoth f = apply f f

I originally thought that if "apply f f" is valid, then logically "applyBoth f" should also be valid. But it seems that type inference results in applyBoth only working for functions "c a a -> c a2 a2".

applyBoth f = apply f f In apply, (f1 :: c a b -> c a2 b2) and (f2 :: c b a -> c b2 a2). So for "apply f f" to typecheck, we must have that a ~ b and a2 ~ b2 (as the above two type signatures must match since f1 = f2). So we must have that (f :: c a a -> c a2 a2).

Is there a way to type "applyBoth" so it works for all functions that would work simply by repeating them twice in "apply"?

I'm not sure what you mean; what else would make sense? Can you provide an example? -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com http://IvanMiljenovic.wordpress.com