What does `someObject.new` do in Java?

Question

In Java, I have just found out that the following code is legal:

KnockKnockServer newServer = new KnockKnockServer();                    
KnockKnockServer.re         


        

Answer 1

It's the way to instantiate a non-static inner class from outside the containing class body, as described in the Oracle docs.

Every inner class instance is associated with an instance of its containing class. When you new an inner class from within its containing class it uses the this instance of the container by default:

public class Foo {
  int val;
  public Foo(int v) { val = v; }

  class Bar {
    public void printVal() {
      // this is the val belonging to our containing instance
      System.out.println(val);
    }
  }

  public Bar createBar() {
    return new Bar(); // equivalent of this.new Bar()
  }
}

But if you want to create an instance of Bar outside Foo, or associate a new instance with a containing instance other than this then you have to use the prefix notation.

Foo f = new Foo(5);
Foo.Bar b = f.new Bar();
b.printVal(); // prints 5

Answer 2

When inner classes were added to Java in version 1.1 of the language they were originally defined as a transformation to 1.0 compatible code. If you look at an example of this transformation, I think it will make it a lot clearer how an inner class actually works.

Consider the code from Ian Roberts' answer:

public class Foo {
  int val;
  public Foo(int v) { val = v; }

  class Bar {
    public void printVal() {
      System.out.println(val);
    }
  }

  public Bar createBar() {
    return new Bar();
  }
}

When transformed to 1.0 compatible code, that inner class Bar would become something like this:

class Foo$Bar {
  private Foo this$0;

  Foo$Bar(Foo outerThis) {
    this.this$0 = outerThis;
  }

  public void printVal() {
    System.out.println(this$0.val);
  }
}

The inner class name is prefixed with the outer class name so as to make it unique. A hidden private this$0 member is added that holds a copy of the outer this. And a hidden constructor is created to initialise that member.

And if you look at the createBar method, it would be transformed into something like this:

public Foo$Bar createBar() {
  return new Foo$Bar(this);
}

So let's see what happens when you execute the following code.

Foo f = new Foo(5);
Foo.Bar b = f.createBar();                               
b.printVal();

First we instantiate an instance of Foo and intialise the val member to 5 (i.e. f.val = 5).

Next we call f.createBar(), which instantiates an instance of Foo$Bar and initialises the this$0 member to the value of this passed in from createBar (i.e. b.this$0 = f).

Finally we call b.printVal() which tries to print b.this$0.val which is f.val which is 5.

Now that was a regular instantiation of an inner class. Let's look at what happens when instantiating Bar from outside Foo.

Foo f = new Foo(5);
Foo.Bar b = f.new Bar();
b.printVal();

Applying our 1.0 transformation again, that second line would become something like this:

Foo$Bar b = new Foo$Bar(f);

This is almost identical to the f.createBar() call. Again we're instantiating an instance of Foo$Bar and initialising the this$0 member to f. So again, b.this$0 = f.

And again when you call b.printVal(), you are printing b.thi$0.val which is f.val which is 5.

The key thing to remember is that the inner class has a hidden member holding a copy of this from the outer class. When you instantiate an inner class from within the outer class, it it implicitly initialised with the current value of this. When you instantiate the inner class from outside the outer class, you explicitly specify which instance of the outer class to use, via the prefix on the new keyword.

Answer 3

Have a look at this example:

public class Test {

    class TestInner{

    }

    public TestInner method(){
        return new TestInner();
    }

    public static void main(String[] args) throws Exception{
        Test t = new Test();
        Test.TestInner ti = t.new TestInner();
    }
}

Using javap we can view instructions generated for this code

Main method:

public static void main(java.lang.String[])   throws java.lang.Exception;
  Code:
   0:   new     #2; //class Test
   3:   dup
   4:   invokespecial   #3; //Method "<init>":()V
   7:   astore_1
   8:   new     #4; //class Test$TestInner
   11:  dup
   12:  aload_1
   13:  dup
   14:  invokevirtual   #5; //Method java/lang/Object.getClass:()Ljava/lang/Class;
   17:  pop
   18:  invokespecial   #6; //Method Test$TestInner."<init>":(LTest;)V
   21:  astore_2
   22:  return
}

Inner class constructor:

Test$TestInner(Test);
  Code:
   0:   aload_0
   1:   aload_1
   2:   putfield        #1; //Field this$0:LTest;
   5:   aload_0
   6:   invokespecial   #2; //Method java/lang/Object."<init>":()V
   9:   return

}

Everything is simple - when invoking TestInner constructor, java passes Test instance as a first argument main:12. Not looking at that TestInner should have a no argument constructor. TestInner in its turn just saves reference to parent object, Test$TestInner:2. When you are invoking inner class constructor from an instance method, reference to parent object is passes automatically, so you do not have to specify it. Actually its passes every time, but when invoking from outside it should be passed explicitly.

t.new TestInner(); - is just a way to specify the first hidden argument to TestInner constructor, not a type

method() is equal to:

public TestInner method(){
    return this.new TestInner();
}

TestInner is equal to:

class TestInner{
    private Test this$0;

    TestInner(Test parent){
        this.this$0 = parent;
    }
}

Answer 4

Think of new receiver as a single token. Kind of like a function name with a space in it.

Of course, the class KnockKnockServer does not literally have a function named new receiver, but I'm guessing the syntax is meant to suggest that. It's meant to look like you're calling a function that creates a new instance of KnockKnockServer.receiver using a particular instance of KnockKnockServer for any accesses to the enclosing class.

Answer 5

Shadowing

If a declaration of a type (such as a member variable or a parameter name) in a particular scope (such as an inner class or a method definition) has the same name as another declaration in the enclosing scope, then the declaration shadows the declaration of the enclosing scope. You cannot refer to a shadowed declaration by its name alone. The following example, ShadowTest, demonstrates this:

public class ShadowTest {

    public int x = 0;

    class FirstLevel {

        public int x = 1;

        void methodInFirstLevel(int x) {
            System.out.println("x = " + x);
            System.out.println("this.x = " + this.x);
            System.out.println("ShadowTest.this.x = " + ShadowTest.this.x);
        }
    }

    public static void main(String... args) {
        ShadowTest st = new ShadowTest();
        ShadowTest.FirstLevel fl = st.new FirstLevel();
        fl.methodInFirstLevel(23);
    }
}

The following is the output of this example:

x = 23
this.x = 1
ShadowTest.this.x = 0

This example defines three variables named x: The member variable of the class ShadowTest, the member variable of the inner class FirstLevel, and the parameter in the method methodInFirstLevel. The variable x defined as a parameter of the method methodInFirstLevel shadows the variable of the inner class FirstLevel. Consequently, when you use the variable x in the method methodInFirstLevel, it refers to the method parameter. To refer to the member variable of the inner class FirstLevel, use the keyword this to represent the enclosing scope:

System.out.println("this.x = " + this.x);

Refer to member variables that enclose larger scopes by the class name to which they belong. For example, the following statement accesses the member variable of the class ShadowTest from the method methodInFirstLevel:

System.out.println("ShadowTest.this.x = " + ShadowTest.this.x);

Refer to the docs