Static and Non-Static initialization blocks in Java

Question

Why does this code print 11 and not 10. Clearly , i++ in the static initialization block is executed.
But , why i-- in the non-static block is not executed.
What's happening here ?

class ClassOne
{
    static int i = 10;

    {
        i--;
    }

}

public class Main extends ClassOne
{
    static
    {
        i++;
    }

    public static void main(String[] args)
    {
        System.out.println(i);

    }
}

k5_ · Accepted Answer · 2019-09-15 12:42:34Z

8

Non-static initialization blocks will be called on instance creation.

You never create a new instance, so the block is not executed.

answered Sep 15, 2019 at 12:42

k5_

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2 Comments

k5_ Over a year ago

Anyway, please don't use instance initialization blocks. Many don't know they even exist, and it is very confusing to find them in code. Just stick to regular constructors or field initialization.

Andy Turner Over a year ago

never say never. There are legitimate (but rare) uses for them.

Ilya Sereb · Accepted Answer · 2019-09-15 17:22:07Z

1

This is specifically related to instance initializer blocks and static initializer block. In the above example, the block with i-- is instance initializer block and will be executed each time a new instance of either Main or ClassOne is created.

Static initializer blocks are executed at the time of class loading. So when Main class is being loaded into memory it's parent class is loaded first, and then the variable is getting loaded into memory as well. Thereafter static block in Main gets executed, resulting in 11 to be printed on the console.

edited Sep 15, 2019 at 17:22

Ilya Sereb

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answered Sep 15, 2019 at 14:14

Ritika Goel

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