Is there a way to specify a subset of type parameters in Scala, inferring the rest?

Question

I have a class which looks like this:

class X[A <: Throwable, B, C](b: B, c: C)

A, B and C can be inferred, so I can just instantiate it with:

val x = new X(3, 4)

which gives me an X[Nothing, Int, Int] - often what I want.

but I sometimes want to specify A to be something other than Nothing (say AssertionError). Is this possible without also specifying B and C. I imagined syntax along the lines of:

val x = new X[AssertionError](3, 4)
val x = new X[AssertionError, _, _](3, 4)
val x = new X[AssertionError,,](3, 4)

but obviously this doesn't work.

Is there some syntax for this, or some way I can achieve the same result?

Answer 1

Here is my solution:

scala> class X[A <: Throwable, B, C](b: B, c: C)
defined class X

scala> class Builder[A <: Throwable] {
     |   def apply[B, C](b: B, c: C) = new X[A,B,C](b,c)
     | }
defined class Builder

scala> def X[A <: Throwable]: Builder[A] = new Builder[A]
X: [A <: Throwable]=> Builder[A]

scala> val x = X[AssertionError](3, 4)
x: X[AssertionError,Int,Int] = X@2fc709