两个接口具有相同的方法名称和不同的返回类型

Question

This question is for understanding interfaces in Java. Here is a very simple example of implementing a interfaces in Java.

interface ParentA {
    void display();
}

interface ParentB {
    int display();
}

class Child implements ParentA, ParentB {
    @Override
    public void display() {
        System.err.println("Child ParentA");
    }
    //ERROR : The return type is incompatible with ParentB.display()

    //So added method with int return type too
    @Override
    public int display() {
        System.err.println("Child ParentB");
    }
}

This case can happen in large Java application where two interface can have method with same name. I thought that since return type is different JVM will know which interface's method we are overriding.

What is the best explanation for this? Does this situation make sense?

Thanks in Advance

Answer 1

Because method with same signature is not allowed, It confuses compiler to detect the exact override-equivalent method from the declared once.

JLS (§8.4.2)

Two methods or constructors, M and N, have the same signature if they have,

• the same name
• the same type parameters (if any) ( §8.4.4 ), and
• after adapting the formal parameter types of N to the the type parameters of M, the same formal parameter types.

It is a compile-time error to declare two methods with override-equivalent signatures in a class.

Answer 2

If you found yourself in this situation and couldn't solve in a cleaner way, you could have one or two inner classes that forward the calls to newly named methods:

class Child {
    private class ParentAImp implements ParentA {
      @Override
      public void display() {
          displayParentA();
      }
    }

    private class ParentBImp implements ParentB {
      @Override
      public int display() {
          return displayParentB(); 
      }
    }

    public ParentA asParentA(){ return new ParentAImp(); }
    public ParentB asParentB(){ return new ParentBImp(); }

    private void displayParentA() {
        System.err.println("Child ParentA");
    }

    private int displayParentB() {
        System.err.println("Child ParentB");
        return 0;
    }
}

Drawback is now to get from Child to interface you have to do:

ParentA parentA = child.asParentA();
ParentB parentB = child.asParentB();

Answer 3

It is not possible to have in the same class two methods with the same signature and different return type.

Definition of signature:

The combination of the method name and the parameter list.

This happens because it is not possible for the jvm two choose between to methods that differs only from the return type. The jvm infact can investigate only the name of the method and the type of parameters of the call.

So your example is not possible in java.

Probably you have to create two classes, one for interface.

Answer 4

To implement overriding you are breaking overloading rules.

The conditions for method overloading

1. The number of parameters is different for the methods.
2. The parameter types are different (like changing a parameter that was a float to an int).

So in your case, you can not have same method name having same method arguments (in your case no arguments) and different return type.

---

If you add a parameter to one of your interface method then your code will compile.

interface ParentA {
    void display();
}

interface ParentB {
    int display(int x);
}

class Child implements ParentA, ParentB {
    @Override
    public void display() {
        System.err.println("Child ParentA");
    }

    @Override
    public int display(int x) {
        System.err.println("Child ParentB");
        return 0;
    }
}

Answer 5

If you think about it from the side of some code calling methods on an instance of Child:

Child child = new Child();
child.display();

Which display method would be called? Seeing as it could be either method, you'd see a compile time error.

An interesting way round this in Java 8 would be to use a default method in one of the interfaces; these come with their own set of rules.