super.a = b instead of super(b)

user_14168597 0
Viewed 58

I'm learning the super keyword and accidentally get this, here's an example :

    public class A {
       double a;
       public A (double a) {
          this.a = a;
       }
    }
    public class B extends A {
       public B (double b) {
          super.a = b;  //***
       }
    }

The usual way to do this as in the tutorials is super(b) to reusing its parent constructor, but what is wrong with super.a = b?

extends super inheritance constructor java
2 Answers

There is no "constructor inheritance" in Java. If the super class doesn't have a no-arg constructor, you must explicitly call its constructor (or one of them if there are several). If you don't, you'll get a compilation error.

user_24227760

When you write your class A like this:

public class A {
    double a;
    public A(double a) {
        this.a = a;
    }
}

you overwrite the default constructor and in the line this.a = a you are accessing instance variable and setting the values and in class B:

public class B extends A {
    public B(double b) {
        super.a = b; // ***
    }
}

you are trying to access instance variables of Class B through the constructor because super.a in here , the super referes to the constructor and its wrong and throwing the Implicit super constructor A() is undefined. Must explicitly invoke another constructor which means: in Class B its looking for a constructor which has no parameter because you overwrite the default constructor of class and it can't recognize it by calling super.a = b so you have to call the super constructor as a function and in the first line of code:

public class B extends A {
    public B(double b) {
        super(b);
        super.a = b; // ***
    }
}
user_55152870
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