[java] Implementing two interfaces in a class with same method. Which interface method is overridden?

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;
  }
}   

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.

As in interface,we are just declaring methods,concrete class which implements these both interfaces understands is that there is only one method(as you described both have same name in return type). so there should not be an issue with it.You will be able to define that method in concrete class.

But when two interface have a method with the same name but different return type and you implement two methods in concrete class:

Please look at below code:

public interface InterfaceA {
  public void print();
}


public interface InterfaceB {
  public int print();
}

public class ClassAB implements InterfaceA, InterfaceB {
  public void print()
  {
    System.out.println("Inside InterfaceA");
  }
  public int print()
  {
    System.out.println("Inside InterfaceB");
    return 5;
  }
}

when compiler gets method "public void print()" it first looks in InterfaceA and it gets it.But still it gives compile time error that return type is not compatible with method of InterfaceB.

So it goes haywire for compiler.

In this way, you will not be able to implement two interface having a method of same name but different return type.

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(){

}

}

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.

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.

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