2

output here is = 5 but not 15 why according to the code value of x is modified with 15

class A {
    int x = 5;

    public int getX() {
        return this.x;
    }

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

}

class B extends A {
    int x = 10;

    public B() {
        this.x = 15;
    }

}

public class Test {
    public static void main(String[] args) {
        B a = new B();
        System.out.println("hello" + a.getX());
    }
}

is it because of the scope of variable

deHaar
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suyash sahu
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  • If what you're saying is really happening, it must be because of the initialization sequence. The fix is to initialize your state from the constructors, where you can visibly control the order in which things happen. – Tim Biegeleisen Sep 05 '18 at 05:39
  • Just go through this https://stackoverflow.com/questions/5630486/what-happens-when-base-and-derived-classes-each-have-variables-with-the-same-nam – vinay chhabra Sep 05 '18 at 05:49

4 Answers4

2

You redeclared x again in B:

class B extends A {
    int x = 10; // <--- here

This creates two fields called x. I will refer to them as A.x and B.x from now on. Note that B.x hides A.x. This means that if the compile time type of the variable is A, you can only access A.x, whereas if the compile time type is B, you can only access B.x.

When you do new B(), A.x gets initialised to 5. B.x gets set to 15 in the constructor. Simple so far.

But then you called getX(). Note that this is a method from A. It does not know the existence of B.x because inside it, the compile time type of this is A. So inside getX, this.x means A.x. This is why 5 is returned.

To output 15, you can just access B.x directly:

System.out.println(a.x);

Alternatively, B can override A.getX and A.setX to return and set B.x:

class B extends A {
    int x = 10;

    public B() {
        this.x = 15;
    }

    @Override
    public int getX() {
        return x;
    }

    @Override
    public void setX(int x) {
        this.x = x;
    }
}
Sweeper
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2

a.getX() calls the get method of super class and thus returns the value of x in class A.

When you add getter in class B you get value of x as 15.

class A {
    int x = 5;

    public int getX() {
        return this.x;
    }

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

}

class B extends A {
    int x = 10;

    public B() {
        this.x = 15;
    }

    public int getX() {
        return this.x;
    }

}

public class Test {
    public static void main(String[] args) {
        B a = new B();
        System.out.println("hello" + a.getX());
    }
}
Pooja Aggarwal
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  • 6
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1

You can't reinitialize variable in subclass which exists in superclass. This way when getting x it will return the x value of superclass instead of value of subclass. So in B class instead of int x = 10 use x = 10

Michał Gryniuk
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1

The getX() is not defined in the class B, so it will refer the super class for the reference of the method, since the reference is found in the super class it will execute the method getX() of super class.

If you add getX() in class B then the same will be executed.

class A {
    int x = 5;

    public int getX() {
        return this.x;
    }

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

}

class B extends A {
    int x = 10;

    public B() {
        this.x = 15;
    }

    public int getX() {
        return this.x;
    }

}

public class Test {
    public static void main(String[] args) {
        B a = new B();
        System.out.println("hello" + a.getX());
    }
}

This results 15.

Sri Krishna V
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