Scala超类是否有可能将此作为子类返回

Question

abstract class SuperClass {
  def method() = this
}

class SubClass1 extends SuperClass {
  def method1() = this
}
class SubClass2 extends SuperClass {
  def method2() = this
}

val obj1 = new SubClass1()
obj1.method1().method() // this is ok
val obj2 = new SubClass2()
obj2.method().method2() // this is not ok, as method returns a SuperClass, 
                        // which has no method named method2

So how to make the method return the subclass type so I can chain the operation in any order.

So far what I can think is to follow something like

abstract class SuperClass[T <: SuperClass[T]]

but I don't know how to continue that.

Answer 1

Something like this should work:

abstract class SuperClass {
  def method(): this.type = this
}

Answer 2

I find the following approach works as well:

abstract class SuperClass[T <: SuperClass[T]] {
  def method(): T = this.asInstanceOf[T]
}

class SubClass extends SuperClass[SubClass] {
  def method1() = this
}

val obj = new SubClass()
obj.method1().method()
obj.method().method1()