scala generic method overriding

Question

I have an abstract class :

abstract class Foo(...){    def bar1(f : Foo) : Boolean    def bar2(f : Foo) : Foo } 

multiple classes extend Foo and override the methods

class FooImpl(...) extends Foo{     override def bar1(f : Foo) : Boolean {         ...     }     override def bar2(f : Foo) : Foo {         ...     } }  

Is it possible, using generics (or something) to make the overriding methods have the parametertype of the subclass implementing it? Like this :

class FooImpl(...) extends Foo{     override def bar1(f : FooImpl) : Boolean {         ...     }     override def bar2(f : FooImpl) : FooImpl {         ...     } } 

I was thinking something along the line of the following, but that didn't seem to work...

abstract class Foo(...){     def bar1[T <: Foo](f : T) : Boolean     def bar2[T <: Foo](f : T) : T }  class FooImpl(...) extends Foo{     override def bar1[FooImpl](f : FooImpl) : Boolean {        ...     }     override def bar2[FooImpl](f : FooImpl) : FooImpl{        ...     } } 

Any help is much appreciated!

Thank you.

Answer 1

abstract class Foo{    type T <: Foo    def bar1(f:T):Boolean    def bar2(f:T):T }  class FooImpl extends Foo{    type T = FooImpl    override def bar1(f:FooImpl) = true    override def bar2(f:FooImpl) = f } 

In this version, different subclasses of Foo all share Foo as a superclass, but to hold the return value of bar2 (or the parameters to bar1 or bar2) in a setting where all you know about your object (let's say it's named obj) is that it's a Foo, you need to use the type obj.T as the type of the variable.

Answer 2

To make Ken Blum's second version a little bit nicer you can use self types:

abstract class Foo[T] { self:T =>    def bar1(f:T):Boolean    def bar2(f:T):T }  class FooImpl extends Foo[FooImpl]{    override def bar1(f:FooImpl) = true    override def bar2(f:FooImpl) = f }