Why does the no args have to be present for the code to work?

Question

On this piece of code I understand it works with either super(t) in Class A or a no args constructor in class B. Below the code uses the no args constructor in class B. I understand that if you take the no args constructor out of class B the code does not work. I'm new to programming and what I'm trying to understand is what is so special about the no args constructor in class B, Why does it have to be present for the code to work? What is the special condition or rule?

public class Test {

    public static void main(String[] args) {

       B b = new B(5);

    }
}

class A extends B {
    public A(int t)  {

        System.out.println("A's constructor is invoked");
    }
}

class B  {

    public B() {

    }

    public B(int k)   {
    System.out.println("B's constructor is invoked");
    }
}

Answer 1

class A extends B {

Compiler ensures while compiling this class; if programmer doesn't write this() or super() explicitely, it will add super() as very first line in every constructor provided in class A.

And compiler also ensures to add no-arg constructor to a class if no other constructor in provided by programmer.

Now, lets say you don't provide no-arg constructor in class B and compiler will also no provide cause you have provided constructor with argument. But, in class A it'll add super() as very first line in constructor. i.e. a call to base class no-arg construtor. Thus, it'll lead to compiler error.

So, as you said, either provide no-arg constructor in class B OR write super(<int>) in class A.