Monday, July 19, 2010

Java Object Initialization Order - Know your JLS!

Recently I came across an interesting problem whose solution eluded me at first glance. Consider these three classes:
package com.ds.test;

public class Upper {
 String upperString;

 public Upper() {
  Initializer.initialize(this);
 }
}
package com.ds.test;

public class Lower extends Upper {

 String lowerString = null;

 public Lower() {
  super();
  System.out.println("Upper:  " + upperString);
  System.out.println("Lower:  " + lowerString);
 }

 public static void main(final String[] args) {
  new Lower();
 }
}
package com.ds.test;
public class Initializer {
 static void initialize(final Upper anUpper) {
  if (anUpper instanceof Lower) {
   Lower lower = (Lower) anUpper;
   lower.lowerString = "lowerInited";
  }
  anUpper.upperString = "upperInited";
 }
}
What output is to be expected from running the Lower class?

In this very reduced example it is much easier to get a view of the whole situation - in reality where this occurred there was a lot more code to distract one's attention...
Anyway, this is what the output looks like:
Upper:  upperInited
Lower:  null;
While the little example uses Strings, the real code of Initializer had a delegate object registered with the equivalent of the Lower class - at least that was the intention. For some reason however did this not work when running the application. Instead, the default path was taken - the one for the delegate object being not set (null).
Now, change the code of Lower slightly:
package com.ds.test;

public class Lower extends Upper {

 String lowerString;

 public Lower() {
  super();
  System.out.println("Upper:  " + upperString);
  System.out.println("Lower:  " + lowerString);
 }

 public static void main(final String[] args) {
  new Lower();
 }
}
The output is now:
Upper:  upperInited
Lower:  lowerInited
Notice the difference in the code?
Yes, the lowerString field is no longer explicitly set to null. Why would this make a difference? Isn't the default value for reference type fields (such as String here) null anyway? Of course, it is. However it turns out that this tiny little change - which apparently would not change the code's behavior in any way - makes this thing fly or not fly.
So what is going on? It becomes clear when looking at the initialization order:
  1. main() calls the Lower constructor.
  2. An instance of Lower is prepared. That means, all fields are created and populated with default values, i. e. null for reference types, false for booleans and so on. At this time, any inline assignments to the fields have not taken place!
  3. The super-constructor is called. This is mandated by the language spec. So, before anything else happens, Upper's constructor is called.
  4. The Upper constructor runs and hands a reference to the freshly created instance to the Initializer.initialize() method.
  5. The Initializer attaches new Strings to both fields. It does so by using a somewhat dirty instanceof check - not a particularly good design pattern, but possible, nevertheless. Once that has happened, both the upperString lowerString references are no longer null.
  6. The Initializer.initialize() call finishes, as does the Upper constructor.
  7. Now it becomes interesting: Construction of the Lower instance continues. Assuming there is no explicit =null assignment in the lowerString field declaration, the Lower constructor resumes execution and prints out the two Strings that are attached to the fields.
    However, if there is an explicit assignment to null, execution has a slightly different flow: Just after the super constructor is done, any variable initializers are executed (see section 12.5 of the Java Language Spec), before the rest of the constructor is run. In this case the String reference that was previously assigned to lowerString is now overwritten with null again! Only then does the rest of the constructor continue execution, now printing lowerString: null.
Apart from being a nice example for why it is handy to be aware of some of the minutiae of object creation (or knowing where to look in the JLS, printed or online) this shows why it is a bad idea to write the Initializer like this. It should not be aware of Upper's subclasses at all! Instead, if for some reason initialization of certain fields cannot be done in the Lower class itself, it will just require its own variant of some sort of initialization helper. In that case, it would really make no difference if you used String lowerString; or String lowerString = null; - just as it should be.

5 comments:

Anonymous said...

Initially thought it was just rereading the uninitialized space declared for lowerString, but I see now that you are right.

Interesting...

Anonymous said...

Guess that only happens in C tho.

Parag Shah said...

Hi Daniel, I am the founder of diycomputerscience.com and am making a matrix of Java competencies.

One of the competencies is understanding how Java initializes objects.

I like the way you have explained the initialization order of Java classes and would like to seek your permission to use this example with a slight modification (am thinking of adding a few static fields and a static initializer).

Can I use your example on my website ?

Thanks

Daniel Schneller said...

Sure. It's not a secret :)

Parag Shah said...

Thanks Daniel. I have created the competency. I have added some static members and a static initializer, but besides that the example is largely unchanged.

You can view the competency at http://diycomputerscience.com/competencies/topic/core-java/competency/understands-the-order-of-initialization-of-members-in-a-class