OCJP的误解

Question

I'm preparing to take OCJP exam and I have a tricky question. I don't know why the correct answer is the next one : "B. 300-300-100-100-100" The question sounds like this:

1.    class Foo {

2.        private int x;

3.        public Foo(int x) {
4.            this.x = x;
5.        }

6.        public void setX(int x) {
7.            this.x = x;
8.        }

9.        public int getX() {
10.            return x;
11.        }
12.    }

13.    public class Gamma {

14.       static Foo fooBar(Foo foo) {
15.            foo = new Foo(100);
16.            return foo;
17.        }

18.        public static void main(String[] args) {
19.            Foo foo = new Foo(300);
20.            System.out.println(foo.getX() + "-");

21.            Foo fooBoo = fooBar(foo);
22.            System.out.println(foo.getX() + "-");
23.            System.out.println(fooBoo.getX() + "-");

24.           foo = fooBar(fooBoo);
25.            System.out.println(foo.getX() + "-");
26.            System.out.println(fooBoo.getX() + "-");
27.        }
28.    }

Frankly speaking I was expected that correct answer should be "A. 300-100-100-100-100" because at line 15 foo reference is changed to a new Foo object which has as instance variable x=100 and I don't know why at line 22 foo reference is to the "old object" with instance variable x=300

Can someone explain me why? Thanks!

Answer 1

Explaining inline :

public static void main(String[] args) {
19.            Foo foo = new Foo(300);
20.            System.out.println(foo.getX() + "-"); // 300

21.            Foo fooBoo = fooBar(foo);   // foo is "unchanged" here
22.            System.out.println(foo.getX() + "-");  // printing foo --> 300
23.            System.out.println(fooBoo.getX() + "-"); // printing fooBoo --> 100

24.           foo = fooBar(fooBoo); // changing foo and setting it to 100
25.            System.out.println(foo.getX() + "-"); // So foo will be --> 100
26.            System.out.println(fooBoo.getX() + "-");// fooBoo is already -->100
27.        }

Answer 2

在第24行上更改了对foo变量的引用。在第15行上没有更改。因为原始foo被静态方法fooBar的本地foo变量隐藏。

Answer 3

This is how objects work in Java

Foo fooBoo = fooBar(foo); // here foo ---> 300

//function fooBar call

foo = new Foo(100); 

This line will create new foo object pointing to 100 so it will look like this

foo (passed to method) ----> 300
foo (created inside method) ---> 100

That is why the value is not changing in that step