Cannot change a variable value

Question

So, I'm making a simple program that is using 2 classes.

The first class contains this:

public class A{
    private int x;

    public A(){
        x = changexvalue(x);
        System.out.println(x); //Check value
    }

    private int changexvalue(int x){
        x = x + 2;
        return x;
    }

    public int getxvalue(){
        return x;
    }
}

The second class contains this:

public class B{

    public static void main(String [] args){
        A a = new A();
        System.out.println(a.getxvalue());
    }
}

So, the problem is this. The first output(in class A) prints 2 but the class B output shows 0(I want the class B output show 2). How is this possible and how can I fix this?

Thanks

Possible duplicate of http://stackoverflow.com/questions/9360663/variable-can-not-be-changed?rq=1 — Dmitry Ginzburg, May 27 '14 at 13:57
`x = changexvalue(x);` where in your method you are doing `x = x + 2;` is confusing. Do you think that you are passing variable `x` or value of `x` to `changexvalue` method? Do you know that Java is always [pass-by-value](http://stackoverflow.com/questions/40480/is-java-pass-by-reference-or-pass-by-value)? — Pshemo, May 27 '14 at 14:02
How can you access `a` in `B.main` when it's not `static`? Did you simplify your code and deleted too much? — Markus Ratzer, May 27 '14 at 14:05
How it's possible that you use a(non static variable) from static main function? — Ostap Maliuvanchuk, May 27 '14 at 14:09
you should make a static. Also Java is case sensitive, there no Public keyword in it. How it even compiles for you? — Ostap Maliuvanchuk, May 27 '14 at 14:35
@user3680043 The code you are showing does not produce the output you describe. Please show your actual code. For instance, your class `A` won't compile, given that you've capitalized `Public`. — GriffeyDog, May 27 '14 at 14:35

score 0 · Answer 1 · answered May 27 '14 at 13:59

0

Use this.x to refer to the class' member variable x. x on its own refers to the local variable.

answered May 27 '14 at 13:59

Bathsheba

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Which local variable are you talking about. There no local variable x in A constructor. – Ostap Maliuvanchuk May 27 '14 at 14:01
To elaborate: in `changexvalue`, the `x` that is being changed/returned (and thus printed when `a` is created) is the local `x` (the parameter), NOT the instance variable `x`. An alternate fix would be to use a different name for the local variable. – Scott Hunter May 27 '14 at 14:01
this.x on the getxvalue() method or the changexvalue method? – user3680043 May 27 '14 at 14:01
@user3680043: Use a different name than just `x`. `y` would suffice. – Engineer2021 May 27 '14 at 14:08
@user3680043: `this.x` would mean the same thing (the member variable) in both, but its only in `changexvalue` that `x` does not mean the same thing as `this.x`. – Scott Hunter May 27 '14 at 14:10
1

It doesn't matter. The method returns the value of the local `x`, _which is then assigned to the instance variable `x` by the constructor._ – GriffeyDog May 27 '14 at 14:37

score 0 · Answer 2 · answered May 27 '14 at 14:03

0

x got shadowed here. You use two variablees with the same name so only the inner variable (the input of the method at this point) is accessable and x the attribute of your class can not be accessed.

private int changexvalue(int x){
    x = x + 2;
    return x;
}

replace it with

private int changexvalue(int y){
    x = y + 2;
    return x;
}

or use this to make clear where you use the x Attribute of the class and where you use the x input variable of your method:

private int changexvalue(int x){
    this.x = x + 2;
    return this.x;
}

answered May 27 '14 at 14:03

Simulant

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or just `return y + 2;`. – Engineer2021 May 27 '14 at 14:09
2

And calling member functions from constructors is bad practice anyway. – Bathsheba May 27 '14 at 14:10
this no the cause, even if x is shadowed in changexvalue it's should be overwritten in A(),because there no local x variable. – Ostap Maliuvanchuk May 27 '14 at 14:11
@Bathsheba why bad practice? – Grim May 27 '14 at 14:12
Someone might override the function in a child class. Then all hell will break loose. In C++ the behaviour is undefined. – Bathsheba May 27 '14 at 14:13
@PeterRader: because you can override the method by an extending class and thereby changing the behavior of the constructor without knowing of the changed behavior. – Simulant May 27 '14 at 14:13
@PeterRader methods are polymorphic so if you override it in derived class constructor of super class will invoked overridden method, which may cause many problems. You are safe only in case of private/final/static methods which are not polymorphic. – Pshemo May 27 '14 at 14:13
1

Not bad prictise if it's final and private – Ostap Maliuvanchuk May 27 '14 at 14:13
It doesn't matter. The method returns the value of the local x, _which is then assigned to the instance variable x by the constructor_. – GriffeyDog May 27 '14 at 14:38

Cannot change a variable value

2 Answers2