Why I cannot increment integer in void method

Question

I am really confuse why in this small example i is still 0:

public static void main(String[] args) {
    int i = 0;
    inc(i);
    System.out.println(i);
}

private static void inc(int i) {
    i++;
}

probably very easy question but I dont see it

Answer 1

Java passes parameters by value.

So the i in the method inc is really just a copy of the "original" i in the main method. You increment that copy, but that has no influence on the original variable i outside.

See this question for a more detailed explanation.

Answer 2

This is because you're using a primitive integer, essentially you're passing the value of i to inc not a reference to i. In this case just return a value from your inc method:

public static void main(String[] args) {
    inc i = 0;
    i = inc(i);
    System.out.println(i);
}

private static int inc(int i) {
    return i++;
}

Of course, if you're passing around objects then you are passing by reference and so you will be able to mutate without returning.