class and interface initialization in java

Question

Below is the code,

class Z{
    static int peekj(){
        return j;
    }
    static int peekk(){
        return k;
    }
    static int i = peekj();
    static int h = peekk();
    static final int j = 1;
    static int k = 1;
}


public class ClassAndInterfaceInitialization {
    public static void main(String[] args) {
        System.out.println(Z.i);
        System.out.println(Z.h);
    }
}

After following the forward reference rule for static initialization, I see the output as:

1
0

After class Z is loaded & linked, In the initialization phase, variable j being final is very firstly initialized with 1. variable k is also initialized with 1.

But the output gives 0 for variable k.

How do I understand this?

Note: Compiler actually replaces value of variable j wherever it is referenced following forward reference rules, unlike k

One more question : System.out.println(Z.peekk()); outputs 1 where as System.out.println(Z.h); outputs 0. How? — Naman Gala
– Naman Gala, Commented Aug 18, 2015 at 5:18
@overexchange Yes All the 'J' occurences are replaced by actual Number 1 . See my answer below. — Oliver
– Oliver, Commented Aug 18, 2015 at 5:53

TheLostMind · Accepted Answer · 2015-08-18 04:58:17Z

6

static final int will make j a compile-time constant. So, its value will be present and passed as part of the byte-code itself. So, j will be 1 when your class is being initialized.But k is is just static int, so its default value will be printed since your static initializer runs before the value of k is initialized.

answered Aug 18, 2015 at 4:58

TheLostMind

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5 Comments

Naman Gala Over a year ago

One more question : System.out.println(Z.peekk()); outputs 1 where as System.out.println(Z.h); outputs 0. How?

saagarjha Over a year ago

Once System.out.println(Z.peekk()) is called, all of the variables are properly initialized (to 1).

Naman Gala Over a year ago

In main method : System.out.println(Z.i); System.out.println(Z.peekk()); System.out.println(Z.h);. Output: 1 1 0

Naman Gala Over a year ago

In main method: System.out.println(Z.peekk()); System.out.println(Z.h);. Output: 1 0. No change in output. Note: Even I change the order of SOP then also the behavior is same.

TheLostMind Over a year ago

@NamanGala - Thats because when h was being initialized, k was 0. But peekk() returns the current value of k when the method is called. You are calling method after the class is initialized. SO, value is 1. got it?

saagarjha · Accepted Answer · 2015-08-18 05:18:45Z

5

You are trying to access k before it is initialized, which is why it is 0. (h is initialized to peekk() before k is defined, so peekk() returns 0, as fields are initialized from the top down.) j is final, so it is execute first; before the declaration of i, thus i gets the value of j or 1.

Here's the appropriate Oracle documentation:

Next, execute either the class variable initializers and static initializers of the class, or the field initializers of the interface, in textual order, as though they were a single block.

edited Aug 18, 2015 at 5:18

answered Aug 18, 2015 at 4:54

saagarjha

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5 Comments

Codebender Over a year ago

The same is happening for j also. But it's giving proper result because of compile-time replacement (as stated in TheLostMind's answer).

saagarjha Over a year ago

@TimBiegeleisen I did it.

Naman Gala Over a year ago

One more question : System.out.println(Z.peekk()); outputs 1 where as System.out.println(Z.h); outputs 0. How?

overexchange Over a year ago

When you say, "trying to access k before it is initialized, which is why it is 0", Are you saying that, there is a separate phase prior to initialization phase, where k is set to 0? Is yes, what is that phase?

saagarjha Over a year ago

Basically. Java will provide default values to everything before everything else, before all other code is run (at compile time, IIRC).

Oliver · Accepted Answer · 2015-08-18 05:47:54Z

3

"j" is a constant so when the code is compiled, j is replaced by the number 1 in the whole program. So if you were an alien and you could read byte code the code would look like this

 static int peekj(){
        return 1;// j is replaced by the Actual Number 1
    }

Also "static variables are initialized when the class is loaded" So when the class loads the following statements are executed in sequential manner

     static int i = peekj(); // Executed First, as peekJ() directly returns 1 so i is equal to 1
        static int h = peekk(); // Executed Second, Now peekk() returns "THE VALUE OF k" which is not set yet i.e default value of int is given 0(zero)
//So h is equal to 0

        static final int j = 1; // This statement doesn't exist as all j's are replaced with actual Number 1 when we compiled java to .class 
        static int k = 1; // Executed last, So now the value of 'k' is being set, after the value of 'h' has already been set to zero

Note: Do this to get output 1 and 1 i.e i = 1 and h =1; as the value of k is being set to 1 first.

  static int k = 1;
 static int i = peekj();
    static int h = peekk();
    static final int j = 1;

answered Aug 18, 2015 at 5:47

Oliver

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2 Comments

overexchange Over a year ago

One supple,What are the defaults values of different primitive data types?

Oliver Over a year ago

@overexchange Click Here docs.oracle.com/javase/tutorial/java/nutsandbolts/… And Scroll down to see the table

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