2 Oct
2011
2 Oct
'11
10:13 a.m.
On Sun, Oct 2, 2011 at 9:00 AM, Felipe Almeida Lessa wrote: On Sun, Oct 2, 2011 at 12:04 AM, Rustom Mody How would you classify a function of type (Int, Int) -> Int -> Int ? It's curried. Uncurried would be: ((Int, Int), Int) -> Int Yes, if you mean that uncurrying it gives the type ((Int, Int), Int) -> Int
But it can also be curried to get the type Int -> Int -> Int -> Int
Does that not make it uncurried as well? Likewise if we have a polymorphic foo: Int -> a and we instantiate a to
Int
-> Int does foo suddenly get curried? Int -> a is both curried and uncurried. Int -> Int -> Int is just curried. --
Felipe.