Implementing two interfaces in a class with same m

Question

Two interfaces with same method names and signatures. But implemented by a single class then how the compiler will identify the which method is for which interface?

Ex:

interface A{
  int f();
}

interface B{
  int f();
}

class Test implements A, B{   
  public static void main(String... args) throws Exception{   

  }

  @Override
  public int f() {  // from which interface A or B
    return 0;
  }
}   

Answer 1

As far as the compiler is concerned, those two methods are identical. There will be one implementation of both.

This isn't a problem if the two methods are effectively identical, in that they should have the same implementation. If they are contractually different (as per the documentation for each interface), you'll be in trouble.

Answer 2

Try implementing the interface as anonymous.

public class MyClass extends MySuperClass implements MyInterface{

MyInterface myInterface = new MyInterface(){

/* Overrided method from interface */
@override
public void method1(){

}

};

/* Overrided method from superclass*/
@override
public void method1(){

}

}

Answer 3

There is nothing to identify. Interfaces only proscribe a method name and signature. If both interfaces have a method of exactly the same name and signature, the implementing class can implement both interface methods with a single concrete method.

However, if the semantic contracts of the two interface method are contradicting, you've pretty much lost; you cannot implement both interfaces in a single class then.

Answer 4

Well if they are both the same it doesn't matter. It implements both of them with a single concrete method per interface method.

Answer 5

This was marked as a duplicate to this question https://stackoverflow.com/questions/24401064/understanding-and-solving-the-diamond-problems-in-java

You need Java 8 to get a multiple inheritance problem, but it is still not a diamon problem as such.

interface A {
    default void hi() { System.out.println("A"); }
}

interface B {
    default void hi() { System.out.println("B"); }
}

class AB implements A, B { // won't compile
}

new AB().hi(); // won't compile.

As JB Nizet comments you can fix this my overriding.

class AB implements A, B {
    public void hi() { A.super.hi(); }
}

However, you don't have a problem with

interface D extends A { }

interface E extends A { }

interface F extends A {
    default void hi() { System.out.println("F"); }
}

class DE implement D, E { }

new DE().hi(); // prints A

class DEF implement D, E, F { }

new DEF().hi(); // prints F as it is closer in the heirarchy than A.

Answer 6

If a type implements two interfaces, and each interface define a method that has identical signature, then in effect there is only one method, and they are not distinguishable. If, say, the two methods have conflicting return types, then it will be a compilation error. This is the general rule of inheritance, method overriding, hiding, and declarations, and applies also to possible conflicts not only between 2 inherited interface methods, but also an interface and a super class method, or even just conflicts due to type erasure of generics.

---

Compatibility example

Here's an example where you have an interface Gift, which has a present() method (as in, presenting gifts), and also an interface Guest, which also has a present() method (as in, the guest is present and not absent).

Presentable johnny is both a Gift and a Guest.

public class InterfaceTest {
    interface Gift  { void present(); }
    interface Guest { void present(); }

    interface Presentable extends Gift, Guest { }

    public static void main(String[] args) {
        Presentable johnny = new Presentable() {
            @Override public void present() {
                System.out.println("Heeeereee's Johnny!!!");
            }
        };
        johnny.present();                     // "Heeeereee's Johnny!!!"

        ((Gift) johnny).present();            // "Heeeereee's Johnny!!!"
        ((Guest) johnny).present();           // "Heeeereee's Johnny!!!"

        Gift johnnyAsGift = (Gift) johnny;
        johnnyAsGift.present();               // "Heeeereee's Johnny!!!"

        Guest johnnyAsGuest = (Guest) johnny;
        johnnyAsGuest.present();              // "Heeeereee's Johnny!!!"
    }
}

The above snippet compiles and runs.

Note that there is only one @Override necessary!!!. This is because Gift.present() and Guest.present() are "@Override-equivalent" (JLS 8.4.2).

Thus, johnny only has one implementation of present(), and it doesn't matter how you treat johnny, whether as a Gift or as a Guest, there is only one method to invoke.

---

Incompatibility example

Here's an example where the two inherited methods are NOT @Override-equivalent:

public class InterfaceTest {
    interface Gift  { void present(); }
    interface Guest { boolean present(); }

    interface Presentable extends Gift, Guest { } // DOES NOT COMPILE!!!
    // "types InterfaceTest.Guest and InterfaceTest.Gift are incompatible;
    //  both define present(), but with unrelated return types"
}

This further reiterates that inheriting members from an interface must obey the general rule of member declarations. Here we have Gift and Guest define present() with incompatible return types: one void the other boolean. For the same reason that you can't an void present() and a boolean present() in one type, this example results in a compilation error.

---

Summary

You can inherit methods that are @Override-equivalent, subject to the usual requirements of method overriding and hiding. Since they ARE @Override-equivalent, effectively there is only one method to implement, and thus there's nothing to distinguish/select from.

The compiler does not have to identify which method is for which interface, because once they are determined to be @Override-equivalent, they're the same method.

Resolving potential incompatibilities may be a tricky task, but that's another issue altogether.

References

• JLS 8.4.2 Method Signature
• JLS 8.4.8 Inheritance, Overriding, and Hiding
• JLS 8.4.8.3 Requirements in Overriding and Hiding
• JLS 8.4.8.4 Inheriting Methods with Override-Equivalent Signatures
  • "It is possible for a class to inherit multiple methods with override-equivalent signatures."