CODEWITHSUNDEEP
scalawildcardtype-parameter
Jean-Philippe Pellet
Jean-Philippe Pellet

Reputation: 59994

Why can't a wildcard type parameter is Scala be bound?

I have a typed pair class:

class TypedPair[T]

and I want to apply a certain function to a heterogeneous sequence of them:

def process[T](entry: TypedPair[T]) = {/* something */}

Why doesn't this work?

def apply(entries: TypedPair[_]*) = entries.foreach(process)

It fails with the error:

error: polymorphic expression cannot be instantiated to expected type;
 found   : [T](TypedPair[T]) => Unit
 required: (TypedPair[_]) => ?
         def apply(entries: TypedPair[_]*) = entries.foreach(process)

I don't recall getting into this problem in Java...

Upvotes: 2

Views: 1870

Answers (2)

Daniel C. Sobral
Daniel C. Sobral

Reputation: 297175

You have declared an existential type:

def apply(entries: TypedPair[_]*) = entries.foreach(process)

is equivalent to

def apply(entries: TypedPair[t] forSome { type t }*) = entries.foreach(process)

I'm not sure if this is what you intended or not.

Upvotes: 0

Debilski
Debilski

Reputation: 67838

The compiler has problems figuring out the anonymous method in this case. When you added the dummy parameter, you also changed the syntax to help the compiler with it, so the following will work:

def apply(entries: TypedPair[_]*) = entries.foreach(process(_))

Upvotes: 3

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