Scala: Why doesn't this compile?

Question

Given:

class Foo[T] { def get: T } class Bar class FooBar extends Foo[Bar] { def get = new Bar } object Baz { def something [T, U <: Foo[T]] (foo : Class[U]): T = foo.newInstance.get }

I should be able to do something like this, right?

Baz.something(classOf[FooBar])

Strangely this is throwing:

inferred type arguments [Nothing,this.FooBar] do not conform to method something's type parameter bounds [T,U <: this.Foo[T]]

Which is weird :S. BTW I'm having this issue while migrating some java code that's equivalent to what I write here and it's working fine.

Foo class is missing an abstract keyword.

pedrofurla
– pedrofurla

2012-10-18 18:59:12 +00:00
Commented Oct 18, 2012 at 18:59 — pedrofurla
– pedrofurla, Commented Oct 18, 2012 at 18:59

Community · Accepted Answer · 2017-05-23 11:56:19Z

You've run into one of the more annoying limitations of Scala's type inference! See this answer for a clear explanation of why the compiler is choking here.

You have a handful of options. Most simply you can just provide the types yourself:

Baz.something[Bar, FooBar](classOf[FooBar])

But that's annoyingly verbose. If you really don't care about U, you can leave it out of the type argument list:

object Baz { def something[T](foo: Class[_ <: Foo[T]]): T = foo.newInstance.get }

Now FooBar will be inferred correctly in your example. You can also use a trick discussed in the answer linked above:

object Baz { def something[T, U <% Foo[T]](foo: Class[U]): T = foo.newInstance.get }

Why this works is a little tricky—the key is that after the view bound is desugared, T no longer appears in U's bound.

Régis Jean-Gilles · Accepted Answer · 2012-10-18 19:03:10Z

It does not compile because T does not appear anywhere in the parameter list and thus cannot be inferred(or rather, it is inferred to Nothing).

You can fix it like this:

def something [T] (foo : Class[_ <: Foo[T]]): T = foo.newInstance.get

pedrofurla · Accepted Answer · 2012-10-18 19:03:25Z

Adding the types explicitly seems to work:

scala> Baz.something[Bar,FooBar](classOf[FooBar]) res1: Bar = Bar@3bb0ff0

Usually, when you get a Nothing where you expect some other type to be inferred means that for some reason the compiler is not able to infer the type in that particular position.

That said, I propose the question: how could we refactor the code so the type gets inferred? @Régis' answer shows the solution.

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