Array of Interface in Java

问题:

I have an interface.

public interface Module {
        void init();
        void actions();
}

What happens when i try to create an array like this?

Module[] instances = new Module[20]

How can i implement this array?

回答1:

yes, it is possible. You need to fill the fields of the array with objects of Type Module

instances[0] = new MyModule();

And MyModule is a class implementing the Module interface. Alternatively you could use anonymous inner classes:

instances[0] = new Module() {
 public void actions() {}
 public void init() {}
};

Does this answer your question?

回答2:

You would need to fill the array with instances of a class(es) that implement that interface.

Module[] instances = new Module[20];
for (int i = 0; i < 20; i++)
{
    instances[i] = new myClassThatImplementsModule();
}

回答3:

You need to create a concrete class type that would implement that interface and use that in your array creation

回答4:

Of course you can create an array whose type is an interface. You just have to put references to concrete instances of that interface into the array, either created with a name or anonymously, before using the elements in it. Below is a simple example which prints hash code of the array object. If you try to use any element, say myArray[0].method1(), you get an NPE.

public class Test {
 public static void main(String[] args) {
     MyInterface[] myArray = new MyInterface[10];
     System.out.println(myArray);
 }
 public interface MyInterface {
     void method1();
     void method2();
 }
}

回答5:

To clarify accepted answer from @Burna, this array can be used to arrange collection of object, but it can never arrange its own interface in the collection, that is to put Module interface in instances reference variable. In JLS Chapter 10.6

Each variable initializer must be assignment-compatible (§5.2) with the array's component type, or a compile-time error occurs.

But you can't use the interface Module in the initialization, because interface can't be instantiated (by definition). Thus you have to implement it first and arrange it in the array.