Java generic method inheritance and override rules

Question

I have an abstract class that has a generic method and I want to override the generic method by substituting specific types for the generic parameter. So in pseudo-code I have the following:

public abstract class GetAndParse {
  public SomeClass var;

  public abstract <T extends AnotherClass> void getAndParse(T... args);
}

public class Implementor extends GetAndParse {
  // some field declarations

  // some method declarations

  @Override
  public <SpecificClass> void getAndParse(SpecificClass... args) {
    // method body making use of args
  }
}

But for some reason I'm not allowed to do this? Am I making some kind of syntax error or is this kind of inheritance and overriding not allowed? Specifically I'm getting an error about @Override because the eclipse IDE keeps reminding me to implement getAndParse.

Here's how I want the above code to work. Somewhere else in my code there is a method that expects instances of objects that implement GetAndParse which specifically means that they have a getAndParse method that I can use. When I call getAndParse on that instance the compiler checks to see whether I have used specific instances of T in the proper way, so in particular T should extend AnotherClass and it should be SpecificClass.

Hest

pseudo code too abstract, need more info – Sibelius

Using public abstract <T extends AnotherClass> void getAndParse(Args... args); makes no sense. What is the type parameter good for? How should the compiler determine its actual value and where should it use it? – Carousal

Could you get us a complete example, which is in fact compilable (apart from the problematic error) and produces the error you mentioned? – Euniceeunuch

@maaartinus: You're right I've changed the methods to make specific use of the generic type. – Hest

Answer 1

What we are having here is two different methods with individual type parameters each.

public abstract <T extends AnotherClass> void getAndParse(Args... args);

This is a method with a type parameter named T, and bounded by AnotherClass, meaning each subtype of AnotherClass is allowed as a type parameter.

public <SpecificClass> void getAndParse(Args... args)

This is a method with a type parameter named SpecificClass, bounded by Object (meaning each type is allowed as a type parameter). Do you really want this?

Is the type parameter used inside Args? I think the problem would be there.

---

The meaning of

public abstract <T extends AnotherClass> void getAndParse(T... args);

is that the caller of the method can decide with which type parameter he wants to call the method, as long as this is some subtype of AnotherClass. This means that in effect the method can be called with any objects of type AnotherClass.

Since the caller can decide the type parameter, you can't in a subclass narrow down the parameter type to SpecificClass - this would not be an implementation of the method, but another method with same name (overloading).

Maybe you want something like this:

public abstract class GetAndParse<T extends AnotherClass> {
  public SomeClass var;

  public abstract void getAndParse(T... args);
}

public class Implementor extends GetAndParse<SpecificClass> {
  // some field declarations

  // some method declarations

  @Override
  public void getAndParse(SpecificClass... args) {
    // method body making use of args
  }
}

Now the getAndParse method implements the parent class' method.

Yamamoto

I don't want a type parameter. I want the compiler to check and make sure that SpecificClass is an extension of AnotherClass. – Hest

Answer 2

You are seeing this problem because of the concept called "Erasure" in Java Generics. Java uses "erasure" to support backward compatibility. i.e Java code which did not use generics.

Erasure Procedure:
The compiler will first do a type checking and then it will remove(erase) all the type parameters as much as possible, and also insert TypeCasting where ever necessary.

example:

public abstract <T extends AnotherClass> void getAndParse(T paramAnotherClass);

will become

public abstract void getAndParse(AnotherClass paramAnotherClass);

In class "Implementor.java",

The code

public <SpecificClass> void getAndParse(T paramAnotherClass)

will become

public void getAndParse(SpecificClass paramAnotherClass){  }

the compiler will see that you have not implemented the abstract method correctly. There is a type mismatch between the abstract method and the implemented method. This is why you are seeing the error.

More details can be found here. http://today.java.net/pub/a/today/2003/12/02/explorations.html

Contraption

Answer 3

You cannot override to specific type T because there is in fact (at the bytecode level if you wish) only one method getAndParse because of type erasure (see other answer):

public abstract void getAndParse(AnotherClass... args); // (1)

For every type of T, the same method is used.

You can overload it (I think):

public void getAndParse(SpecificClass... args); // (2)

but this will not a different method from (1) ant it will not be called by generic code:

T x = whatever;
object.getAndParse(x); // Calls (1) even if T is derived from SpecificClass

Vulva

Answer 4

No, it's not valid. What would happen if someone with a GetAndParse reference called it with a different class extending AnotherClass?

Vaudois

I don't understand why this is a problem. I have an implementor that extends GetAndParse and fixes the type of T to SpecficClass which the compiler checks to make sure extends AnotherClass then in my code when I have an instance of Implementor and call getAndParse with a type that doesn't match SpecificClass I should get an error. – Hest

@davidk01: The contract provided by GetAndParse say that everyone implementing it must do so for any type extending AnotherClass. Given GetAndParse foo = new Implementor(); you must able to call foo.getAndParse for any extension of AnotherClass - If the compiler didn't enforce this, you could just as well use Object, the point of generics is to ensure compile-time type safety – Vaudois

Answer 5

That becomes a nonsense when someone has a reference to type GetAndParse and tries to call the getAndParse method. If Cat and Dog extend AnotherClass. I should expect to be able to call GetAndParse#getAndParse with either a Cat or a Dog. But the implementation has tried to restrict it and make it less compatible!

Breakout

That is the reason that what you want to do is impossible, I apologize if I framed it in a way you don't like. Compile time generics are not capable of enforcing that for you. How is the compiler to know what subclass will be there at runtime? It's entirely possible the implementor is returned by a factory interface and comes from a different library you're not even compiling against! Narrowing of the type parameter cannot be checked for safety at compile time. – Breakout

Answer 6

Static method can't override

class Vehicle{
static void park(int location){
    System.out.println("Vehicle parking..");
}}

class Car extends  Vehicle{
@Override //error
void park(int location) { //error
    System.out.println("Car Parking..");
}}

Private method can't override

class Vehicle{
private void park(int location){
    System.out.println("Vehicle parking..");
}
void callPark(){
    park(100);
}}

class Car extends  Vehicle{
//@Override
void park(int location) {
    System.out.println("Car Parking..");
}}

class Demo {
public static void main(String[] args) {
    Vehicle v1=new Car();
   v1.callPark();
}}

Final method can't override

class Vehicle{
final void park(int location){
    System.out.println("Vehicle parking..");
}}

class Car extends  Vehicle{
//@Override
void park(int location) { //error
    System.out.println("Car Parking..");
}}

Halliard

Can you be more descriptive by what you mean with your answer? I could guess that you're listing ways where Override doesn't work. But can you give an introduction why this has anything to do with this question? Please edit your answer to make it better. – Derive