0

Lets see a small snippet of java code

class Foo {
  int i = 0;
  int j;
  public Foo(int j){
     this.j = j;
  }
}

Above example shows two ways of initializing variables in Java. But my question is which variable first get initialized? The variable outside the constructor or the variable inside the constructor? When I instantiate the above class Foo foo = new Foo(5), I know that the constructor of the class get called which implies variable j get initialized first. Can anyone make me clear about the ordering.

Improve this question
5
  • please comment if the question deserves downvote Commented Jan 15, 2014 at 4:53
  • there are multiple questions on SO for this. Also u can always try code and confirm Commented Jan 15, 2014 at 4:53
  • Actually.. i and j are both initialized at the same time.. After initialization, you are changing the value of j from 0 (default) to something else.. Commented Jan 15, 2014 at 4:56
  • @TheLostMind i am not talking about the default initialization. My query is focused on initialization after and before the constructor Commented Jan 15, 2014 at 5:00
  • @BibekSubedi - Check my answer.. Hope it answers your question... Commented Jan 15, 2014 at 5:05

5 Answers 5

Reset to default
2

Precedence

In your case the int j happens first and defaults to 0, then gets reassigned to 5 when the constructor is called on to create a new instance.

j only gets re-assigned when the constructor runs. Instance members get initialized first when you assign them something outside the constructor.

Order of Execution

Each line of code gets executed in the order it appears. The declarations happen before the constructor always, in the order they are listed in the code.

Deterministic and Predictable

You should only initialize instance members in a single place, inside a single constructor.

Relying on defaults leads to hard to track down bugs, and makes testing a nightmare, but a instance member that is unassigned will stand out like a sore thumb to the IDE, the compiler and at runtime. Unfortunately for primitives like int they default to 0 which might not be what you want/need.

A better design is:

 private final int j;

 public Foo(final int j) { this.j = j; }

This keeps the j from getting assigned anything on initialization and you never have to worry about it changing.

Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

1

Let's say you have this class

public class Test
{
    int i = 4;

    public Test()
    {
        System.out.println(i);
    }  
}

When you create an instance of Test, the output will be 4. This means that the initialization goes first. Then the constructor is called. Because if the constructor were first and the initialization of i second, you won't get 4 as output.

Improve this answer

2 Comments

hmm, logical answer. Interesting but not sure
@BibekSubedi - Christian is right.. constructor is used for initialization.. In java variables are initialized with default values when object is created, so in constructor we perform 2nd round of initialization..
0 
int i = 0;

i is an instance variable and hence will get default value which is 0. So you don't need to assign 0 to it explicitly.

Coming to your question

 But my question is which variable first get initialized?

Then the answer would be i is initialized before j. Because if you see byte codes of your snippet Object is created first which will initialize your instance variables and then constructor is called.

Improve this answer

Comments

0

When ever you create Object creation, First declaration of variable are being done.After that only variables are defined. In your case first "j" variable will declared first i.e. class variable.After that the value will be assigned to "j" in constructor of a class.

Improve this answer

Comments

0

Ordering :-

  1. First of all static initilization blocks are executed.(If any)
  2. Then execution starts from main method.

for your question I think i and j will be initailized together as i=0; and j as 0 (with int default value.)

Because in constructor you are just changing the value.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.