Java Instance Initialization Block and instance variables

Question

In context to my previous question Java classes and static blocks what if I changed my code from static block and variables to normal Instance Initialization Block and instance variables. Now how would the code be executed?

class extra3 {
    public static void main(String string[]) {
        Hello123 h = new Hello123();
        System.out.println(h.a);
    }
}
class Hello123 {
    {
        a=20;
    }
    int a=158;
}

Here I am getting output as 158. I am not able to understand the reason here.And the other code being this:

class extra3 {
    public static void main(String string[]) {
        Hello123 h = new Hello123();
        System.out.println(h.a);
    }
}
class Hello123 {
    int a=158;
    {
        a=20;
    }
}

Here the output is 20 which is acceptable because when object is created Instance block is executed first. But why is the output in first code coming as 158?

What I can see, Code executed serially and that is how it should be. — Subhrajyoti Majumder
– Subhrajyoti Majumder, Commented Jul 23, 2014 at 8:47

Abimaran Kugathasan · Accepted Answer · 2014-07-23 08:46:19Z

5

This is the order of initialization

Set fields to default initial values (0, false, null)
Call the constructor for the object (but don't execute the body of the constructor yet)
Invoke the constructor of the superclass
Initialize fields using initializers and initialization blocks
Execute the body of the constructor

So, when you are initializing fields, both inline initializer (a = 158) and initialization blocks (a = 20) are executed in the order as they have defined.

So in the first case, inline initializer executed after the initialization block, you get 158 as result.

In the second case, initialization block got executed after inline initializer.

answered Jul 23, 2014 at 8:46

Abimaran Kugathasan

32.7k11 gold badges81 silver badges109 bronze badges

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

3 Comments

Shashank Agarwal Over a year ago

I still have 1 doubt. Why did 'a' compile when surely 'a' hasn't been declared when it is set in the instance block

Abimaran Kugathasan Over a year ago

@ShashankAgarwal : I don't understand your question

javadg Over a year ago

@ShashankAgarwal - 'a' is compiled properly because when constructor is called, the instance variable are set to default values in the step 1 itself. See the "accepted answer" ordering. 'a' is set to 0 , then to 20 and then to 158 in example 1

Suresh Atta · Accepted Answer · 2014-07-23 08:45:46Z

1

Order matters.

Initialization's and static blocks executes based on the order they placed in source code. That's the reason.

answered Jul 23, 2014 at 8:45

Suresh Atta

122k38 gold badges206 silver badges315 bronze badges

Comments

Prasad Kharkar · Accepted Answer · 2014-07-23 08:46:25Z

1

Static blocks are executed in the order they are declared in code. This article will help you understand the order of execution of static and non static initlaization blocks

answered Jul 23, 2014 at 8:46

Prasad Kharkar

13.6k5 gold badges42 silver badges57 bronze badges

Collectives™ on Stack Overflow

Java Instance Initialization Block and instance variables

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related