How can I pass an Integer class correctly by reference?

Question

I am hoping that someone can clarify what is happening here for me. I dug around in the integer class for a bit but because integer is overriding the + opporator I could not figure out what was gong wrong. My problem is with this line:

Integer i = 0;
i = i + 1;  // <- I think that this is somehow creating a new object!

Here is my reasoning: I know that java is pass by value (or pass by value of reference), so I think that in the following example the integer object should be incremented each time.

public class PassByReference {

    public static Integer inc(Integer i) {
        i = i+1;    // I think that this must be **sneakally** creating a new integer...  
        System.out.println("Inc: "+i);
        return i;
    }

    public static void main(String[] args) {
        Integer integer = new Integer(0);
        for (int i =0; i<10; i++){
            inc(integer);
            System.out.println("main: "+integer);
        }
    }
}

This is my expected output:

Inc: 1
main: 1
Inc: 2
main: 2
Inc: 3
main: 3
Inc: 4
main: 4
Inc: 5
main: 5
Inc: 6
main: 6
...

This is the actual output.

Inc: 1
main: 0
Inc: 1
main: 0
Inc: 1
main: 0
...

Why is it behaving like this?

Answer 1

There are two problems:

1. Java is pass by value, not by reference. Changing the reference inside a method won't be reflected into the passed-in reference in the calling method.
2. Integer is immutable. There's no such method like Integer#set(i). You could otherwise just make use of it.

To get it to work, you need to reassign the return value of the inc() method.

integer = inc(integer);

---

To learn a bit more about passing by value, here's another example:

public static void main(String... args) {
    String[] strings = new String[] { "foo", "bar" };
    changeReference(strings);
    System.out.println(Arrays.toString(strings)); // still [foo, bar]
    changeValue(strings);
    System.out.println(Arrays.toString(strings)); // [foo, foo]
}
public static void changeReference(String[] strings) {
    strings = new String[] { "foo", "foo" };
}
public static void changeValue(String[] strings) {
    strings[1] = "foo";
}

Answer 2

I think it is the autoboxing that is throwing you off.

This part of your code:

   public static Integer inc(Integer i) {
        i = i+1;    // I think that this must be **sneakally** creating a new integer...  
        System.out.println("Inc: "+i);
        return i;
    }

Really boils down to code that looks like:

  public static Integer inc(Integer i) {
        i = new Integer(i) + new Integer(1);      
        System.out.println("Inc: "+i);
        return i;
    }

Which of course.. will not changes the reference passed in.

You could fix it with something like this

  public static void main(String[] args) {
        Integer integer = new Integer(0);
        for (int i =0; i<10; i++){
            integer = inc(integer);
            System.out.println("main: "+integer);
        }
    }

Answer 3

What you are seeing here is not an overloaded + oparator, but autoboxing behaviour. The Integer class is immutable and your code:

Integer i = 0;
i = i + 1;  

is seen by the compiler (after the autoboxing) as:

Integer i = Integer.valueOf(0);
i = Integer.valueOf(i.intValue() + 1);  

so you are correct in your conclusion that the Integer instance is changed, but not sneakily - it is consistent with the Java language definition :-)

Answer 4

If you change your inc() function to this

 public static Integer inc(Integer i) {
      Integer iParam = i;
      i = i+1;    // I think that this must be **sneakally** creating a new integer...  
      System.out.println(i == iParam);
      return i;
  }

then you will see that it always prints "false". That means that the addition creates a new instance of Integer and stores it in the local variable i ("local", because i is actually a copy of the reference that was passed), leaving the variable of the calling method untouched.

Integer is an immutable class, meaning that you cannot change it's value but must obtain a new instance. In this case you don't have to do it manually like this:

i = new Integer(i+1); //actually, you would use Integer.valueOf(i.intValue()+1);

instead, it is done by autoboxing.

Answer 5

The Integer is immutable. You can wrap int in your custom wrapper class.

class WrapInt{
    int value;
}

WrapInt theInt = new WrapInt();

inc(theInt);
System.out.println("main: "+theInt.value);

Answer 6

You are correct here:

Integer i = 0;
i = i + 1;  // <- I think that this is somehow creating a new object!

First: Integer is immutable. I think the documentation of the Integer class is missing this.

Second: the Integer class is not overriding the + operator, there is autounboxing and autoboxinginvolved at that line (In older versions of Java you would get an error on the above line).
When you write i + 1 the compiler first converts the Integer to an (primitive) int for performing the addition. Next, doing i = <some int> the compiler converts from int to an (new) Integer.
So + is actually being applied to primitive ints.