Java: Why does this swap method not work? [duplicate]

Answer 1

Integer are immutable - you can't change their values. The swapping that occurs inside the swap function is to the references, not the values.

You would need to return both references in an array to achieve what you want

static Integer[] swap(Integer a, Integer b) {
   return new Integer[]{b, a};
}

public static void main(String[] args) {
   Integer a = 1;
   Integer b = 2;

   Integer[] intArray = swap(a, b);

   a = intArray[0];
   b = intArray[1];

   System.out.println("a=" + a + " b=" + b);
} 

If Integer had a setValue method, you could do something like this.

static void swap(Integer a, Integer b) {
   int temp = a.intValue();
   a.setValue(b.intValue());
   b.setValue(temp);
}

But it doesn't - so to achieve what you want, return an array.

Answer 2

This doesn't have anything to do with immutability of integers; it has to do with the fact that Java is Pass-by-Value, Dammit! (Not annoyed, just the title of the article :p )

To sum up: You can't really make a swap method in Java. You just have to do the swap yourself, wherever you need it; which is just three lines of code anyways, so shouldn't be that much of a problem :)

    Thing tmp = a;
    a = b;
    b = tmp;

Answer 3

Everything in Java is passed by value and the values of variables are always primitives or references to object.

Answer 4

Java code:

class swap {

    int n1;
    int n2;
    int n3;

    void valueSwap() {
        n3 = n1;
        n1 = n2;
        n2 = n3;
    }

    public static void main(String[] arguments) {

        Swap trial = new Swap();
        trial.n1 = 2;
        trial.n2 = 3;

        System.out.println("trial.n1 = " + trial.n1);
        System.out.println("trial.n2 = " + trial.n2);

        trial.valueSwap();

        System.out.println("trial.n1 = " + trial.n1);
        System.out.println("trial.n2 = " + trial.n2);

    }
}

Output:

trial.n1 = 2
trial.n2 = 3
trial.n1 = 3
trial.n2 = 2