Multiple parameter closure argument type not inferred

Question

I have a piece of code that I can\'t get to behave in the way I\'d like. I have a class defined in the following way (stripped down for this):

class Behaviou         


        

Answer 1

One of the ways to fix this is to define multiple argument lists. So your map1 method would be defined like this:

def map1[U, V](behaviour: Behaviour[U])(func: (T, U) => V): Behaviour[V] = ...

and you can use it like this:

beh.map1(beh2)((a, b) => a + b)
beh.map1(beh2)(_ + _)

I'm not completely sure why type inference does not work in your case, but I believe that it has something to do with usage of U type parameter. You are using it twice - for the first and second argument. It's probably too complicated for compiler to figure it out. In case of 2 argument lists, U would be inferred during first argument list compilation, and the second argument list will use already inferred type.

Answer 2

See this scala-debate thread for a discussion of what's going on here. The problem is that Scala's type inference happens per parameter list, not per parameter.

As Josh Suereth notes in that thread, there's a good reason for the current approach. If Scala had per-parameter type inference, the compiler couldn't infer an upper bound across types in the same parameter list. Consider the following:

trait X
class Y extends X
class Z extends X

val y = new Y
val z = new Z

def f[A](a: A, b: A): (A, A) = (a, b)
def g[A](a: A)(b: A): (A, A) = (a, b)

f(y, z) works exactly as we'd expect, but g(y)(z) gives a type mismatch, since by the time the compiler gets to the second argument list it's already chosen Y as the type for A.