When does the instance variable get initialized? Is it after the constructor block is done or before it?
Consider this example:
public abstract class Parent {
public Parent(){
System.out.println("Parent Constructor");
init();
}
public void init(){
System.out.println("parent Init()");
}
}
public class Child extends Parent {
private Integer attribute1;
private Integer attribute2 = null;
public Child(){
super();
System.out.println("Child Constructor");
}
public void init(){
System.out.println("Child init()");
super.init();
attribute1 = new Integer(100);
attribute2 = new Integer(200);
}
public void print(){
System.out.println("attribute 1 : " +attribute1);
System.out.println("attribute 2 : " +attribute2);
}
}
public class Tester {
public static void main(String[] args) {
Parent c = new Child();
((Child)c).print();
}
}
Parent Constructor
Child init()
parent Init()
Child Constructor
attribute 1 : 100
attribute 2 : null
When are the memory for the atribute 1 & 2 allocated in the heap ?
Curious to know why is attribute 2 is NULL ?
Are there any design flaws?
When the memory for the atribute 1 & 2 are allocated in the heap ?
The memory for the object as a whole is allocated when the new
operator is invoked, before the java.lang.Object
constructor is entered. Memory is allocated for individual Integer
instances in init
, but there is no point when memory is allocated for individual properties -- only whole objects.
Curious to know why is attribute 2 is NULL ?
The init
method is called in the super constructor, so attribute2
is assigned new Integer(200)
, and then the subclass constructor is invoked which applies property initializers in the order they appear in the source code. This line
private Integer attribute2 = null;
overwrites the value assigned by init()
to null
.
If you add a call to
System.out.println("attribute 2 : " +attribute2);
right after your call to super();
then this will become apparent.
Are there any design flaws?
Calling sub-class methods before the base class has finished initializing is dangerous. The sub-class might rely on its base-class's invariants to protect its own invariants, and if the base-class constructor has not completed, then its invariants may not hold.
This is also likely to confuse C++ programmers and the like who would expect a call to init
from the base class to invoke the base class's version since C++ rewrites the vtable pointer as constructors are entered.
See The Java Language Specification for all the gory details.