Overloading base method in derived class

Question

So I was playing with C# to see if it matched C++ behavior from this post: http://herbsutter.com/2013/05/22/gotw-5-solution-overriding-virtual-functions/ when I came across this very strange behavior:

public class BaseClass
{
    public virtual void Foo(int i)
    {
        Console.WriteLine("Called Foo(int): " + i);
    }

    public void Foo(string i)
    {
        Console.WriteLine("Called Foo(string): " + i);
    }
}

public class DerivedClass : BaseClass
{
    public void Foo(double i)
    {
        Console.WriteLine("Called Foo(double): " + i);
    }
}

public class OverriddenDerivedClass : BaseClass
{
    public override void Foo(int i)
    {
        base.Foo(i);
    }

    public void Foo(double i)
    {
        Console.WriteLine("Called Foo(double): " + i);
    }
}

class Program
{
    static void Main(string[] args)
    {
        DerivedClass derived = new DerivedClass();
        OverriddenDerivedClass overridedDerived = new OverriddenDerivedClass();

        int i = 1;
        double d = 2.0;
        string s = "hi";

        derived.Foo(i);
        derived.Foo(d);
        derived.Foo(s);

        overridedDerived.Foo(i);
        overridedDerived.Foo(d);
        overridedDerived.Foo(s);
    }
}

Output

Called Foo(double): 1
Called Foo(double): 2
Called Foo(string): hi
Called Foo(double): 1
Called Foo(double): 2
Called Foo(string): hi

So apparently it favors the implicitly converted int to double over the more specific Foo(int) from the base class. Or does it hide the Foo(int) from the base class? But then: why isn't Foo(string) hidden? Feels very inconsistent... It also does not matter if I override Foo(int) or not; the outcome is the same. Can anyone explain what's going on here?

(Yes I know that it is bad practice to overload base methods in a derived class - Liskov and all - but I still wouldn't expect that Foo(int) in OverriddenDerivedClass isn't called?!)

Answer 1

To explain how it works for the OverriddenDerivedClass example:

Have a look at the C# spec for member lookup here: http://msdn.microsoft.com/en-us/library/aa691331%28VS.71%29.aspx

That defines how the lookup is done.

In particular, look at this part:

First, the set of all accessible (Section 3.5) members named N declared in T and the base types (Section 7.3.1) of T is constructed. Declarations that include an override modifier are excluded from the set.

In your case, N is Foo(). Because of Declarations that include an override modifier are excluded from the set then the override Foo(int i) is excluded from the set.

Therefore, only the non-overridden Foo(double i) remains, and thus it is the one that is called.

That is how it works for the OverriddenDerivedClass example, but this is not an explanation for the DerivedClass example.

To explain that, look at this part of the spec:

Next, members that are hidden by other members are removed from the set.

The Foo(double i) in DerivedClass is hiding the Foo(int i) from the base class, so it is removed from the set.

The tricky thing here is the part that says:

All methods with the same signature as M declared in a base type of S are removed from the set.

You might say "But wait! Foo(double i) doesn't have the same signature as Foo(int i), so it shouldn't be removed from the set!".

However, because there is an implicit conversion from int to double, it is considered to have the same signature, so Foo(int i) is removed from the set.