Overloading a method based on a type parameter

Question

I'd like to have a method that can be called with or without a type parameter, and return a different value for each. Here's some obviously simplified code:

object Foo {
  def apply() = "Hello"
  def apply[T]() = 1
}

Calling it with a type parameter is fine:

scala> Foo[String]()
res1: Int = 1

But calling it without a type parameter doesn't work:

scala> Foo()
<console>:9: error: ambiguous reference to overloaded definition,
both method apply in object Foo of type [T]()Int
and  method apply in object Foo of type ()java.lang.String
match argument types ()
       Foo()

It's not a runtime problem, so adding an implicit dummy parameter doesn't help. Nor does having a restricted parameter (Foo[Unit]()). Is there any way of doing this that isn't ambiguous to the compiler?

blackgreen · Accepted Answer · 2024-05-08 22:12:39Z

2

You are effectively overloading on the return type. While neither Scala nor Java let you normally let you do so in this case it happens.

Foo() : String will work in this case but it remains questionable though if overloading on the return type is desirable.

edited May 8, 2024 at 22:12

blackgreen♦

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answered Oct 14, 2011 at 14:17

Moritz

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1 Comment

Marcus Downing Over a year ago

You're right, Foo():String works. That's not really desirable though - the whole idea is to simply the calling code. What I really want to say is "I'm calling the version of this argument that takes zero type parameters, not the version that takes one type parameter".

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