Why does order of implementing Interfaces (with default methods) matter in Java 8?

Question

As we all know, multiple interfaces can implemented in Java. Does the order of their implementation matter? I mean, is implementing B, C is same as C, B in Java 8? My tests show order does matter - but can anyone explain the logic behind this?

public interface A {
    public default void display() {
        System.out.println("Display from A");
    }
}

public interface B extends A {
    public default void display() {
        System.out.println("Display from B");
    }
}

public interface C extends A {
    public void display();
}

public interface D extends B, C {

}

The above code works fine. If I change the order B, C to C, B, it will give an error: The default method display() inherited from B conflicts with another method inherited from C.

public interface D extends C, B {

}

Edit

I am using Eclipse(Mars). JDK jdk1.8.0_51. JRE jre1.8.0_60.

Am I missing something? If I understood correctly, you said you get an error when using one order and you don't when using a different order, which means the order makes a difference, which means you answered your own question. — ikegami, Sep 19 '15 at 02:41
Yes, I am getting the difference. I just want to know if there is any reason behind this? — Pradip Kharbuja, Sep 19 '15 at 02:43
You should get the same error regardless of the order (meaning, in both cases). Please check again! — Nir Alfasi, Sep 19 '15 at 02:57
http://stackoverflow.com/questions/31617301/what-are-the-rules-to-handle-homonym-inherited-methods — user2864740, Sep 19 '15 at 03:02

dkatzel · Answer 1 · 2015-09-19T03:49:07.343

8

There should be an error message either way. When I use jdk 1.8.0_31 I get the following error no matter the order of interfaces:

The default method display() inherited from A.B conflicts with another method inherited from A.C

The solution is to either override display() in D even to just tell the compiler which super class's implementation to use:

public interface D extends B, C {
    default void display(){
        B.super.display();
    }
}

Or remake display() abstract

interface D extends B, C {
    public void display();
}

If you are indeed getting this error using a higher version than me, then it might be worth submitting a bug report?

edited Sep 19 '15 at 03:49

answered Sep 19 '15 at 03:38

dkatzel

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D can also re-assert `display()` as abstract. – ZhongYu Sep 19 '15 at 03:42
Error(it is the correct behavior) you see is same in 1.8.0_40. Order makes no difference. – Jayan Sep 19 '15 at 03:48

Stephan Herrmann · Answer 2 · 2015-10-29T12:09:36.707

results from javac (all versions 1.8.0_x):

error: interface D inherits abstract and default for display() from types B and C

resuls from ecj 4.4:

The default method display() inherited from B conflicts with another method inherited from C

resuls from ecj >= 4.4.1:

NO ERROR

Still ecj >= 4.4.1 correctly reports the error if order in D's extends clause is changed.

I conclude that this is a bug in Eclipse, which got introduced in 4.4.1. I've filed bug 477891 to follow up.

Edit: bug 477891 has been fixed in Milestone 3 towards Eclipse 4.6 (GA: June 2016).

Why does order of implementing Interfaces (with default methods) matter in Java 8?

2 Answers2