CareerCup

for example
public interface A {

public void show();
}

public interface B {

public void show();
}

public class Test implements A, B {

public static void main(String[] args) {

A a;
B b;
Test t = new Test();
a = t;
a.show();
b = t;
b.show();

}

@Override
public void show() {
System.out.println("this method is from class Test:");
System.out.println(this.getClass().getInterfaces()[0]);
System.out.println(this.getClass().getInterfaces()[1]);

}

}

op:-
this method is from class Test:
interface A
interface B
this method is from class Test:
interface A
interface B

It will belong to a particular interface based on its child class runtime polymorphism.
If the object instance is created using

IA ia = new TestUtils();
ia.print();

Then its belongs to A interface.

//IA.java

public interface IA {
public void print();
}


//IB.java

public interface IB {
public void print();
}

//TestUtils.java

public class TestUtils implements IA, IB {

@Override
public void print() {
System.out.println("print");
}

public static void main(String[] args) {
IA ia = new TestUtils();
IB ib = new TestUtils();
ia.print();// IA'a print method
ib.print();// IB's print method
}
}

Well actually, it would be of both.

As long as they have the same signatures the compiler will consider them as one method. In case that one of them has a different return type it would result in compile error.

Like virtual method in c++, for multiple inheritance for child class only one method will be appear. same thing is used by compiler here also without using any virtual keyword.
so for class A only one method has to be implemented.

No . problem it works fine when you override its method but when signature of any methods differ then compile time error

It depends on the method on interface B and C. If interface B and C have the same exact method, then it would not be distinguishable. However, if the method for interface B and C is like this,

void someMethod(some parameter)
boolean someMethod(some parameter)

Then it depends what method class A is specifically looking for.