Any way to _not_ call superclass constructor in Ja

Question

If I have a class:

class A {
   public A() { }
}

and another

class B extends A {
   public B() { }
}

is there any way to get B.B() not to call A.A()?

Answer 1

There is absolutely no way to do this in Java; it would break the language specification.

JLS 12 Execution / 12.5 Creation of New Class Instances

Just before a reference to the newly created object is returned as the result, the indicated constructor is processed to initialize the new object using the following procedure:

1. Assign the arguments for the constructor [...]
2. If this constructor begins with an explicit constructor invocation of another constructor in the same class (using this), then [...]
3. This constructor does not begin with an explicit constructor invocation of another constructor in the same class (using this). If this constructor is for a class other than Object, then this constructor will begin with an explicit or implicit invocation of a superclass constructor (using super).
4. Execute the instance initializers and instance variable initializers for this class [...]
5. Execute the rest of the body of this constructor [...]

Answer 2

The closest you can achieve to the desired behaviour is to delegate initialisation normally performed in the constructor to a template method, which you then override in your subclass implementation. For example:

public class A {
  protected Writer writer;

  public A() {
    init();
  }

  protected void init() {
    writer = new FileWriter(new File("foo.txt"));
  }
}

public class B extends A {
  protected void init() {
    writer = new PaperbackWriter();
  }
}

However, as other people have noted this can typically indicate a problem with your design and I typically prefer the composition approach in this scenario; for example in the above code you could define the constructor to accept a Writer implementation as a parameter.

Answer 3

I think the only way to do it is messing up with the byte-code.
I'm not sure if the Classloader or the JVM checks if super() is being called, but, as Bozho wrote, you probably would end with inconsistent objects when doing so.

Answer 4

No and if you could, your derived object wouldn't really be the object it's deriving from now would it? The is-a principle would be violated. So if you really need it, then polymorphism isn't what you're after.

Answer 5

I had a similar requirement where I needed my child class NOT to go through the super class' constructor, and I wanted the rest of the benefits of the super class. Since super class is also mine, here's what I did.

class SuperClass {
    protected SuperClass() {
        init();
    }

    // Added for classes (like ChildClassNew) who do not want the init to be invoked.
    protected SuperClass(boolean doInit) {
        if (doInit)
            init();
    }

    //
}

class ChildClass1 extends SuperClass {
    ChildClass1() {
        // This calls default constructor of super class even without calling super() explicitly.
        // ...
    }
    // ....
}

class ChildClass2 extends SuperClass {
    ChildClass2() {
        // This calls default constructor of super class even without calling super() explicitly.
        // ...
    }
    // ....
}

class ChildClassNew extends SuperClass {
    ChildClassNew() {
        /*
         * This is where I didn't want the super class' constructor to 
         * be invoked, because I didn't want the SuperClass' init() to be invoked.
         * So I added overloaded the SuperClass' constructor where it diesn;t call init().
         * And call the overloaded SuperClass' constructor from this new ChildClassNew.
         */
        super(false);
        // 
        // ...
    }
    // ....
}

Answer 6

If you don't want to call the superclass constructor, there is something else wrong with your object model.