Question

In the example I was expecting to get Third as output since class Third has an exact signature match to the Print call, but the output is Second. Why that happens?

class First
{
    public virtual void Print(string x)
    {
        Console.WriteLine("First");
    }
}

class Second : First
{
    public void Print(object x)
    {
        Console.WriteLine("Second");
    }
}

class Third : Second
{
    public override void Print(string x)
    {
        Console.WriteLine("Third");
    }
}

class Program
{
    static void Main(string[] args)
    {
        string sss = "lalala";
        Third t = new Third();    
        t.Print(sss);

        Console.ReadLine();
    }
}

Answer 1

See here:

…the set of candidates for a method invocation does not include
methods marked override (Section 7.3), and methods in a base class are
not candidates if any method in a derived class is applicable (Section
7.5.5.1).

Thus, overriden method in the Third is not applicable and Second.Print(object) hides base class implementation and becomes single candidate.