Question

My code looks sort of like this, but this is a simplified version:

class A:

public class A{
    public void testArgs(A a){
        System.out.println("A");
    }
    public void test(){
        System.out.println("A");
    }
}

class B:

public class B extends A{
    public void testArgs(B a){
        System.out.println("B");
    }
    public void test(){
        System.out.println("B");
    }
}

class Main:

public class Main{
    public static void main(String[] args){
        a(new B()).testArgs(new B()); // prints A
        (new B()).testArgs(new B());  // prints B
        a(new B()).test();            // prints B
    }
    public static A a(B b){
        return b;
    }
}

Why does a(new B()).testArgs(new B()) print A not B?
Is there some sort of way to workaround/fix this?

edit:
Clarification:

What I really want is the superclass method to be run when it is called with an A, and the subclass method to be run when testArgs is called with a B.

Casting also isn’t an option because in the actual code, unlike here, I don’t know whether the result of the method call is actually B or not.

edit:
Solution:

Thanks everyone for your answers. Thanks for the clarification on overriding. I used this to implement the desired behavior.
For anyone who has a similar problem in the future:
Change class B to

public class B extends A{
    public void testArgs(A a){       // Corrected overriding, thanks
        if(a instanceof B)           // Check if it is an instance of B
            System.out.println("B"); // Do whatever
        else                         // Otherwise
            super.testArgs(a);       // Call superclass method
    }
    public void test(){
        System.out.println("B");
    }
}

Answer 1

The two testArgs functions are different. One takes an A and the other takes a B. Thus, the B version doesnt override the A version. Since a(new B()) is of type A and B extends A, it is the A version that will run.

There is a “workaround”:

public class B extends A{
public void testArgs(A a){ // <-- note how this is A a, not B a 
    System.out.println("B");
}
public void test(){
    System.out.println("B");
}
}

Then you will see B for all 3 cases (because B‘s testArgs will override A‘s testArgs)