Are fields initialized before constructor code is run in Java?

Question

Can anyone explain the output of following program? I thought constructors are initialized before instance variables. So I was expecting the output to be "XZYY".

class X {     Y b = new Y();      X() {         System.out.print("X");     } }  class Y {     Y() {         System.out.print("Y");     } }  public class Z extends X {     Y y = new Y();      Z() {         System.out.print("Z");     }      public static void main(String[] args) {         new Z();     } } 

Answer 1

The correct order of initialisation is:

1. Static variable initialisers and static initialisation blocks, in textual order, if the class hasn't been previously initialised.
2. The super() call in the constructor, whether explicit or implicit.
3. Instance variable initialisers and instance initialisation blocks, in textual order.
4. Remaining body of constructor after super().

See sections §2.17.5-6 of the Java Virtual Machine Specification.

Answer 2

If you look at the decompiled version of the class file

class X {     Y b;      X() {         b = new Y();         System.out.print("X");     } }  class Y {     Y() {         System.out.print("Y");     } }  public class Z extends X {      Y y;      Z() {         y = new Y();         System.out.print("Z");     }      public static void main(String args[]) {         new Z();     } } 

You can find that the instance variable y is moved inside the constructor, so the execution sequence is as follows

1. Call the constructor of Z
2. It triggers the default constructor of X
3. First line of X constructor new Y() is called.
4. Print Y
5. Print X
6. Call the first line in constructor Z new Y()
7. Print Y
8. Print Z

All the instance variables are initialized by using constructor statements.