What will be the output of the program?class Super{ public int i = 0; public Super(String text) /* Line 4 */ { i = 1; } } class Sub extends Super{ public Sub(String text) { i = 2; } public static void main(String args[]) { Sub sub = new Sub("Hello"); System.out.println(sub.i); } }012Compilation fails.
Question
What will be the output of the program?class Super{ public int i = 0; public Super(String text) /* Line 4 */ { i = 1; } } class Sub extends Super{ public Sub(String text) { i = 2; } public static void main(String args[]) { Sub sub = new Sub("Hello"); System.out.println(sub.i); } }012Compilation fails.
Solution
The program will not compile. The reason is that in Java, the constructor of the superclass (Super) is called before the constructor of the subclass (Sub). If the superclass does not have a default constructor (a constructor with no arguments), the subclass must explicitly call one of the constructors of the superclass using the "super" keyword.
In this case, the Super class does not have a default constructor, and the Sub class does not call the existing constructor of the Super class. Therefore, the Java compiler will throw an error.
To fix this, you can add a call to the superclass's constructor in the Sub class's constructor, like this:
public Sub(String text)
{
super(text);
i = 2;
}
With this modification, the program will compile and the output will be "2".
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