CODEWITHSUNDEEP
java
levartsang
levartsang

Reputation: 19

What happend will subClass and SuperClass have same attributes in Java?

public class Test {

    public static void main(String[] args) {
        B b = new B();
        b.test();
    }

}

class A {
    String aString = "a";
    
    void test() {
        System.out.println(aString);
    }
}

class B extends A {
    String aString = "b";
}

why the cant input out "b"? i just think b.test() will call method of superClass, so b will user superClass's attribute.

Upvotes: 0

Views: 1110

Answers (4)

tevemadar
tevemadar

Reputation: 13225

class A {
    String aString = "a";
    
    void test() {
        System.out.println(aString);
    }
}

class B extends A {
    String aString = "b";
}

Creates 2 aString variables for B. While A will not know about B extending it, B can always access "things" from its parent class A, using the super keyword. So when you are in B, you can use aString and super.aString and they will refer the 2 different variables:

class Ideone {
    public static void main (String[] args) {
        A a=new A();
        B b=new B();
        a.test();
        b.test();
        b.suptest();
    }
    
    static class A {
        String aString="A.aString set from A";
        void test() {
            System.out.println("A.test(): "+aString);
        }
    }
    static class B extends A {
        String aString="B.aString set from B";
        {
            super.aString="A.aString set from B";
        }
        void test() {
            System.out.println("B.test(): "+aString);
        }
        void suptest() {
            System.out.print("B.suptest() calling super.test(): ");
            super.test();
        }
    }
}

You can try it on IdeOne, produces output

A.test(): A.aString set from A
B.test(): B.aString set from B
B.suptest() calling super.test(): A.test(): A.aString set from B

Where the first two lines show nothing fancy, A.aString and B.aString both contain their initial value.
But the third output line shows that super.test() call in B really "ends" in A.test(), and that the initializer block in B really altered the inherited aString field.

(Side note: the static class magic relates to the example being contained in a single file, it doesn't affect the inheritance-part)

Upvotes: 1

ControlAltDel
ControlAltDel

Reputation: 35106

The variable in b is hiding the variable in a. You can fix it by removing the type in b

class B extends A {
    public B () {
      aString = "b";
    }
}

Upvotes: 1

Adriel Araujo
Adriel Araujo

Reputation: 1

You are correct.

Since Class A and the method test() are not abstract and the fact that the method test() was not overrided by its subClass B, the return will be exactly how was specified in Class A even if you call it from a B instance.

Upvotes: 0

John Bollinger
John Bollinger

Reputation: 181734

why the cant input out "b"? i just think b.test() will call method of superClass, so b will user superClass's attribute.

Java class and instance variables can be inherited, but they are not virtual. That is, all variable identifiers appearing in Java code are resolved at compile time -- there is no dynamic resolution for them.

Thus, the appearance of identifier aString in method A.test() is resolved at compile time to the aString attribute of class A. This is the attribute that all invocations of that method will access, regardless of the actual class of the object on which it is invoked. If you want to use the attribute of class B when test() is invoked on an instance of that class then provide getter methods in classes A and B and have A.test() obtain the string via those.

Upvotes: 4

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