Java - Parameter Passing Confusion

Question

"Java is pass-by-value"
-Java Language Specification

However, there are things that confuses me. Please take a glimpse of my example

public class Main {

    public static void main(String[] args) {

        StringBuilder myBuilder = new StringBuilder("Michael");

        editString(myBuilder);

        System.out.println(myBuilder);

    }

    public static void editString(StringBuilder x){
        x.append(" Ardan");
    }

}

Output:

Michael Ardan

And this example:

public class Main {

    public static void main(String[] args) {

        int myInt = 10;

        editInt(myInt);

        System.out.println(myInt);

    }

    public static void editInt(int x){
        x ++;
    }
}

Output:

I tried reading other articles and it all says that Java is always pass-by-value. I've done some test scenario. Comparing both example to each other made me think that Java's Objects are pass-by-reference and primitive types are pass-by-value. However, if you tried to replace the int primitive type into Integer Object, the result would be the same. I would love if somebody explain these two examples here.

In Java Objects are always pass-by-reference whereas variables are pass-by-value. Also Strings are immutable objects. — H-Patel, Sep 05 '13 at 02:51
@H-Patel every single article, even JLS says that Java is strictly pass-by-value regardless of datatype. — Michael 'Maik' Ardan, Sep 05 '13 at 02:52
@Thilo It was a good question. I also tried reading the answers and made my scenarios. I was actually asking about what's happening behind my examples. — Michael 'Maik' Ardan, Sep 05 '13 at 03:00

Thilo · Accepted Answer · 2013-09-05T02:56:16.020

Strictly speaking, Java is pass-by-value.

However, what gets passed by value is a "pointer" to an object. So you will be able to access and modify the state of that object from inside the called method (and those changes are visible to everyone else who has a "pointer" to the same object).

Pass-by-reference would mean that you can change where the variable on the calling side points to. That you cannot do in Java.

For example

  public static void editString(StringBuilder x){
     x = new StringBuilder("Foo");
  }

has no effect outside of the method. In pass-by-reference, you would actually be assigning that new StringBuilder back into the calling method variable.

score -2 · Answer 2 · answered Sep 05 '13 at 04:58

you are right , java object always pass-by-reference, but,you can see the source code of wrapper class like 'Integer','String','Short' and so on.they are 'value' are definded as final .it can not be changed .so ,the question 'if you tried to replace the int primitive type into Integer Object, the result would be the same.' which related with it.when you replace with Integer Object and change the value in you method. the Object changes. so, you can see the result.

Java - Parameter Passing Confusion

2 Answers2