Java: pass-by-value references in methods

Question

I read a few articles and similar question in Stackoverflow; however, I did not quite get an answer to my question. Here is the code:

public class CoinFlipping {

    Random random = new Random();

    Boolean head = null;

    public void flip(Boolean b){
        b = random.nextBoolean();
    //      head = b; 
    }

    public static void main(String [] args){
        CoinFlipping cf = new CoinFlipping();
        cf.flip(cf.head);
        System.out.println("Head: "+cf.head);
    }

    }

I refer to this arcticle: Is Java "pass-by-reference" or "pass-by-value"? I can understand why that piece of code behaves as it does. When we call new dog() we basically create a new Dog object. I get it. However, the example that I provided seems to be confusing a bit. I am not creating an object in this case (aren't I?). I am changing a value. Then why in this case when I print results, I get null and not true or false?

Answer 1

The parameter b is passed by value, even though b is itself a reference. Thus, when you call flip, the flip method treats b as a local variable that is a copy of what you pass into it. Changing this copy doesn't change cf.head.

If b were a reference to a mutable object, you could change the object b refers to, and that change would be seen by the rest of the program after flip returns. But a change to b itself is a change to a copy, and any such change will not be seen by the rest of the program. (Boolean is not mutable.)