Question

Today I came up with an interesting problem. I noticed that the following code:

class A
{
    public A()
    {
        Print();
    }
    public virtual void Print()
    {
        Console.WriteLine("Print in A");
    }
}

class B : A
{
    public B()
    {
        Print();
    }

    public override void Print()
    {
        Console.WriteLine("Print in B");
    }
}

class Program
{
    static void Main(string[] args)
    {
        A a = new B();
    }
}

Prints

Print in B
Print in B

I want to know why does it print the “Print in B” twice.

Answer 1

I want to know why does it print the “Print in B” twice.

You’re calling a virtual method twice, on the same object. The object is an instance of B even during A‘s constructor, and so the overridden method will be called. (I believe that in C++, the object only “becomes” an instance of the subclass after the base class constructor has executed, as far as polymorphism is concerned.)

Note that this means that overridden methods called from a constructor will be executed before the derived class’s constructor body has had a chance to execute. This is dangerous. You should almost never call abstract or virtual methods from a constructor, for precisely this reason.

EDIT: Note that when you don’t provide another constructor call to “chain” to using either : this(...) or : base(...) in the constructor declaration, it’s equivalent to using : base(). So B‘s constructor is equivalent to:

public B() : base()
{
    Print();
}

For more on constructor chaining, see my article on the topic.