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Thinking in Java
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Behavior of polymorphic methods
inside constructors

The hierarchy of constructor calls brings up an interesting dilemma. What happens if you’re inside a constructor and you call a dynamically-bound method of the object being constructed? Inside an ordinary method, you can imagine what will happen: The dynamically-bound call is resolved at run time, because the object cannot know whether it belongs to the class that the method is in or some class derived from it. For consistency, you might think this is what should happen inside constructors.

This is not exactly the case. If you call a dynamically-bound method inside a constructor, the overridden definition for that method is used. However, the effect can be rather unexpected and can conceal some difficult-to-find bugs.

Conceptually, the constructor’s job is to bring the object into existence (which is hardly an ordinary feat). Inside any constructor, the entire object might be only partially formed—you can know only that the base-class objects have been initialized, but you cannot know which classes are inherited from you. A dynamically bound method call, however, reaches “outward” into the inheritance hierarchy. It calls a method in a derived class. If you do this inside a constructor, you call a method that might manipulate members that haven’t been initialized yet—a sure recipe for disaster.

You can see the problem in the following example:

// Constructors and polymorphism
// don't produce what you might expect.
import com.bruceeckel.simpletest.*;

abstract class Glyph {
  abstract void draw();
  Glyph() {
    System.out.println("Glyph() before draw()");
    System.out.println("Glyph() after draw()");

class RoundGlyph extends Glyph {
  private int radius = 1;
  RoundGlyph(int r) {
    radius = r;
      "RoundGlyph.RoundGlyph(), radius = " + radius);
  void draw() {
      "RoundGlyph.draw(), radius = " + radius);

public class PolyConstructors {
  private static Test monitor = new Test();
  public static void main(String[] args) {
    new RoundGlyph(5);
    monitor.expect(new String[] {
      "Glyph() before draw()",
      "RoundGlyph.draw(), radius = 0",
      "Glyph() after draw()",
      "RoundGlyph.RoundGlyph(), radius = 5"
} ///:~

In Glyph, the draw( ) method is abstract, so it is designed to be overridden. Indeed, you are forced to override it in RoundGlyph. But the Glyph constructor calls this method, and the call ends up in RoundGlyph.draw( ), which would seem to be the intent. But if you look at the output, you can see that when Glyph’s constructor calls draw( ), the value of radius isn’t even the default initial value 1. It’s 0. This would probably result in either a dot or nothing at all being drawn on the screen, and you’d be left staring, trying to figure out why the program won’t work.

The order of initialization described in the earlier section isn’t quite complete, and that’s the key to solving the mystery. The actual process of initialization is:

  1. The storage allocated for the object is initialized to binary zero before anything else happens. The base-class constructors are called as described previously. At this point, the overridden draw( ) method is called (yes, before the RoundGlyph constructor is called), which discovers a radius value of zero, due to Step 1.
  2. Member initializers are called in the order of declaration. The body of the derived-class constructor is called.

    On the other hand, you should be pretty horrified at the outcome of this program. You’ve done a perfectly logical thing, and yet the behavior is mysteriously wrong, with no complaints from the compiler. (C++ produces more rational behavior in this situation.) Bugs like this could easily be buried and take a long time to discover.

    As a result, a good guideline for constructors is, “Do as little as possible to set the object into a good state, and if you can possibly avoid it, don’t call any methods.” The only safe methods to call inside a constructor are those that are final in the base class. (This also applies to private methods, which are automatically final.) These cannot be overridden and thus cannot produce this kind of surprise.
    Thinking in Java
    Prev Contents / Index Next

   Reproduced courtesy of Bruce Eckel, MindView, Inc. Design by Interspire