14

I was going through C# Brainteasers (http://www.yoda.arachsys.com/csharp/teasers.html) and came across one question: what should be the output of this code?

class Base
{
    public virtual void Foo(int x)
    {
        Console.WriteLine ("Base.Foo(int)");
    }
}

class Derived : Base
{
    public override void Foo(int x)
    {
        Console.WriteLine ("Derived.Foo(int)");
    }

    public void Foo(object o)
    {
        Console.WriteLine ("Derived.Foo(object)");
    }
}

class Test
{
    static void Main()
    {
        Derived d = new Derived();
        int i = 10;
        d.Foo(i);  // it prints ("Derived.Foo(object)"
    }
}

But if I change the code to 

class Derived 
{
    public void Foo(int x)
    {
        Console.WriteLine("Derived.Foo(int)");
    }

    public void Foo(object o)
    {
        Console.WriteLine("Derived.Foo(object)");
    }
}

class Program
{
    static void Main(string[] args)
    {
        Derived d = new Derived();
        int i = 10;
        d.Foo(i); // prints  Derived.Foo(int)");

        Console.ReadKey();
    }
}

I want to why the output is getting changed when we are inheriting vs not inheriting; why is method overloading behaving differently in these two cases?

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3 Answers 3

Reset to default
19

As I specified in the answers page:

Derived.Foo(object) is printed - when choosing an overload, if there are any compatible methods declared in a derived class, all signatures declared in the base class are ignored - even if they're overridden in the same derived class!

In other words, the compiler looks at methods which are freshly-declared in the most derived class (based on the compile-time type of the expression) and sees if any are applicable. If they are, it uses the "best" one available. If none is applicable, it tries the base class, and so on. An overridden method doesn't count as being declared in the derived class.

See sections 7.4.3 and 7.5.5.1 of the C# 3 spec for more details.

Now as for exactly why it's specified like that - I don't know. It makes sense to me that methods declared in the derived class take precedence over those declared in the base class, as otherwise you run into the "brittle base class" problem - adding a method in the base class could change the meaning of code using the derived class. However, if the derived class is overriding the method declared in the base class, it's clearly aware of it, so that element of brittleness doesn't apply.

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2 Comments

forgive my ignorance , but why it is considering method with "object" type parameter as the best one , I think int is more close (like what is happening in 2 nd case)
@Wondering: Yes, int is better - but the int overload is already overriding the method in the base class, so it isn't being considered: it doesn't count as being "freshly declared" in the derived class. That's the whole point of all of this explanation :)
1

It revolves around scope. In the first program, void Foo(int i) belongs to class Base. class Derived merely redefines its behavior.

Foo(int i) is ignored because it's being "borrowed" and redefined via inheritance from class Base. if Foo(object o) did not exist, then Foo(int i) would be used.

In the second program, void Foo(int i) is called because it properly belongs to class Derived (i.e. it's not being borrowed and overriden through inheritance) and has the best signature fit.

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Comments

0

It's because the methods signatures in the derived class are matched first before considering the base class signatures. Therefore, when you try:

d.Foo(i);

It attempts to match the method signature against your current class (not the base class). It finds an acceptable match Foo(object) and doesn't further consider the base class method signatures.

It's all about the compiler finding a matching method signature, and the search order it uses to find these signatures.

-Doug

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3 Comments

The surprising part is that it doesn't include the overriding method in the derived class though. That's still a "method signature in the derived class" but it doesn't count as being declared in the derived class.
"It finds an acceptable match Foo(object)" but , Foo(int )is also acceptable match
Yes, but as Jon notes, Foo(int) isn't declared in the derived class (even though it's overridden, which doesn't count).

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