Parameter type inference in scala

Question

Given a function with two type parameters, is it possible in scala to pass type parameter for a single type, for example just provide type A in the example as type B can be inferred by the compiler from function f.

def foo[A, B](f: A => B): B = {
    f(null.asInstanceOf[A])
}

Right now, I found only two solutions.

Solution 1 (standard usage): Call foo and specify both types foo[String, Int]( e => 1) but the definition of Int is redundant

Solution 2: Change the definition of the function to

def foo[A, B](useType: A => Unit)(f: A => B): B = {
    f(null.asInstanceOf[A])
}

and use it with

def use[T](t: T) = {}

val res: Int = foo(use[String]) { a => 1 }

It works, but it does not seems very beautiful to use a function to provide the type to the compiler.

Is there any way to specify just type A for a function taking two type parameter ?

Dmytro Mitin · Accepted Answer · 2017-10-16T12:05:59.423

Generally no, type parameters can't be partially applied in Scala. Neither in methods nor in classes. Some wrappers can be used:

  def myMethod[A, B]() = ???

  def myMethod1[A0, W <: Wrapper { type A = A0 }]() = ???

  trait Wrapper {
    type A
    type B
  }

Maybe with some path-dependent types:

def myMethod1[A0, W <: Wrapper { type A = A0 }](w: W)(): w.B = ???

And in case of not methods but classes there are approaches with type lambdas and type members (type members can be partially applied: Wrapper { type A = A0 } for trait Wrapper { type A; type B } is an existential type like Wrapper[A0, _] for trait Wrapper[A, B]).

Partially applying type parameters

Thanks for your answer, I will look deeper in this direction, and see if I can use class instead of function. — Pierre-Nicolas Mougel, Oct 16 '17 at 12:05

Parameter type inference in scala

1 Answers1