CODEWITHSUNDEEP
java
Robin
Robin

Reputation: 563

What is Interface.super

I have recently used default methods in java. In its implementation I found 

public interface DefaultMethod {
    default String showMyName(String name){
        return "Hai "+name;

    }
}


public class DefaultMethodMainImpl implements DefaultMethod{

    @Override
    public String showMyName(String name){
        return DefaultMethod.super.showMyName(name);

    }
}

My question is in DefaultMethod.super where super will call it have no super class except Object? what super will return?

Upvotes: 23

Views: 18111

Answers (5)

Gerold Broser
Gerold Broser

Reputation: 14762

JLS 3.9.: super is a keyword.

JLS 15.12.1. Compile-Time Step 1: Determine Class or Interface to Search

  • If the form is TypeName . super . [TypeArguments] Identifier, then:

    • ...

    • Otherwise, TypeName denotes the interface to be searched, I.

            Let T be the type declaration immediately enclosing the method invocation.             It is a compile-time error if I is not a direct superinterface of T, [...]

            | ... |

            | To support invocation of default methods in superinterfaces,
             the TypeName may also refer to a direct superinterface
             of the current class or interface, and the target is that superinterface. |

N.B.: JLS 15.8.3. this

When used as a primary expression, the keyword this denotes a value that is a reference to the object for which the [...] default method was invoked, [...]

[...]

The type of this is the [...] interface type T within which the keyword this occurs.

Upvotes: 0

Markus Mitterauer
Markus Mitterauer

Reputation: 1610

If you use super in a class it usually refers to the ancestor of that class (either the extended class or Object).

In the case of:

  • an overriden default method of an interface
  • or implementing two or more interfaces each with a default method with the same signature

you have to specify the specific interface which default implementation you want to invoke, hence:

<Interface>.super.<method>();

See also The diamond problem.

Upvotes: 42

Yuriy N.
Yuriy N.

Reputation: 6087

Consider this code:

public interface One {
    default void method() {
        System.out.println("method from One");
    }
}

interface Two extends One {
    default void method() {
        System.out.println("method from Two");
    }
}

class Three implements Two {
    
    public void method() {
        Two.super.method();
    }
        
    public static void main(String[] args) {
        new Three().method();
    }
}

Output: method from Two

Two.super.method() is just a clumsy syntax to avoid the diamond problem explained in another answer.
As you can see from the program output, super not really means "calling to ancestor" as it used to be in Java.

Upvotes: 2

shalitha anuradha
shalitha anuradha

Reputation: 119

The main reason is that there could be more than one interface with a super level having a default method with the same method signature. The java compiler needs to understand what Interface is referring to here.

See the below example.

public interface ABC {
    default String showMyName(String name) {
        return "Hai " + name;

    }
}

public interface CDE {
    default String showMyName(String name) {
        return "Hey.........!!! " + name;

    }
}

public class DefaultMethodMainImpl implements ABC, CDE {

    @Override
    public String showMyName(String name) {
        System.out.println(ABC.super.showMyName(name));
        System.out.println(CDE.super.showMyName(name));
        return ABC.super.showMyName(name);
    }
}

Upvotes: 2

Najib Tounsi
Najib Tounsi

Reputation: 131

<Interface>.<method>(); would consider <method> as static, and this is not the case. Hence the use of key word super which is a reference variable that is used to refer a "parent" object.

Upvotes: 13

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