0

I have read that " a variable is shadowed if there is another variable with the same name that is closer in scope". I found this Point class with a constructor as an example:

public class Point {
    public int x = 0;
    public int y = 0;

    public Point(int x, int y) {
        x = x;
        y = y;
    }
}

Then I created an object of the Point class in the CreateObjectDemo class below and printed the value of the variable x.

public class CreateObjectDemo {
    public static void main(String[] args) {

    Point originOne = new Point(23, 94);

    System.out.println(originOne.x);

    }

}

After running the compiler, it prints 0. But why doesn't it print 23? I thought that "x = x" in the constructor would be like "23 = 23". Did I misunderstand the definition of shadowing variables?

  • No, what it means is "the local `x` that's a parameter to `Point` is hiding the class member `x`." So you are hiding one variable that is named the same as a variable in an outer scope. – Joe May 02 '14 at 16:25
  • You are right in that the x (23) gets value 23, but this value 23 is not stored in the class. You simply send it to the consctructor and forget it there. When you after creating a point want to print originOne.x, the value of that class variable gets printed (0) – Sara S May 02 '14 at 16:27

2 Answers2

5

I thought that "x = x" in the constructor would be like "23 = 23".

Within the constructor, the meaning of the simple name x is always just the parameter. So the assignment x = x in the constructor takes the value of the x parameter and assigning it to the x parameter as well. The instance variable is never touched. (It's not clear what you mean by 23 = 23;, so I can't tell whether or not that's accurate.) Basically, this is a no-op and some IDEs will give you a warning about it.

To force it to copy to the instance variable, you want:

this.x = x;

(And likewise for y, of course.)

Jon Skeet
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  • But isn't the value of the parameter `x` equal to `23` after I created the object? If I print out `x` that takes the value of the `x` parameter(23), why does it still print out 0? –  May 02 '14 at 16:49
  • Nvm my comment, I get it now :) ty. –  May 02 '14 at 17:09
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0 is the default value for int type variables, and as the warning says, you're shadowing the variables in the constructor so the object variable is never assigned to.

public Point(int x, int y) {
    x = x; //assign value of local x to local x
    y = y; //assign value of local y to local y
}

You need to use this keyword to refer to the object variables x and y instead of the local variables:

public Point(int x, int y) {
    this.x = x; //assign value of local x to object variable x
    this.y = y; //assign value of local y to object variable y
}
kviiri
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