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.

Answer 1

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?