Question

For example I have this:

class A{
    private int mine = 0; //Some field...
    public A(int a){mine+=a; /*Some operation1...*/}
}


class B extends A{
    private int mine = 0; //Some field...
    public B(int a){mine-=a; /*Some operation2...*/}
}

and I get:

error: constructor A in class A cannot be applied to given types;
    public B(int a){}
    required: int
    found: no arguments
    reason: actual and formal argument lists differ in length
    1 errors

I don’t understand the error? What is telling me to do?

Though, the code works if Constructor of “A” has no arguments.
But I need to do the Operation1 (aka mine+=a;), so I need A’s arguments, but then I fail.

I’m closed in this magic circle. What shall I do?

Answer 1

The first instruction of every constructor is always to invoke one of its superclass constructor. If you don’t do it explicitely, the compiler inserts this instraction for you. The constructor

 public B(int a) {
     mine-=a; 
     /*Some operation2...*/
 }

is thus equivalent to

public B(int a) {
    super(); // invoke the no-arg super constructor
    mine-=a; 
    /*Some operation2...*/
}

Since A doesn’t have a no-arg constructor, the compilation fails. In this case, you must explicitely invoke one of the super constructors:

public B(int a) {
    super(a);
    mine-=a; 
    /*Some operation2...*/
}