Passing arguments to java methods

Question

i Haver been reading the java web page on this, yet am still miffed about what happening. heres the code
   class A {
        public int x;
        A(int x) {
            this.x = x;
        public String toString() {
            return Integer.toString(x);
    public class CallDemo2 {
        static void f(A arg1) {
            arg1 = null;
        public static void main(String args[]) {
            A arg1 = new A(5);
            f(arg1);
            System.out.println("arg1 = " + arg1);
    }The above code prints out 5, i think i understand that because of the local method can just be seen to be intitalling the value. The calling method in the main also something like the value gets passed back not the memory location so therefore doesnt effect the value 10.
Now the below code is what confuses me
class A {
        public int x;
        public A(int x) {
            this.x = x;
        public String toString() {
            return Integer.toString(x);
    public class CallDemo3 {
        static void f(A arg1) {
            arg1.x = 10;
        public static void main(String args[]) {
            A arg1 = new A(5);
            f(arg1);// why does 10 get printed out and not 5 ?
            System.out.println("arg1 = " + arg1);
    }

Answer 1

u have manged to explain how to get to each result, but am not sure
that you understad it correctly, it is different to the way i am
understanding it, thats why im posting here... i hope byt the end of
today i will understand whats going on that bit of code i posted. I
hope someone like pervel can post and explain. Pervel is a reliable
person...sorry cant be of any more help however thsi is the page i am
readingWell, THANKS A LOT. I wont bother to help you again, then. I assure you that my explanation is totally right, but if you dont trust me, well, then I wont help you again.