C# Calling an overriden method within the superclass

Question

Hey there. I just picked up C# to learn game programming with XNA, but I also have some experience in Java.

Here's my code, in essence:

public class A
{
    public Rectangle getRectangle()
    {
        return [some rectangle];
    }

    public bool collision(A other)
    {
        Rectangle rect1 = getRectangle();
        Rectangle rect2 = other.getRectangle();
        return rect1.Intersects(rect2);
    }
}

public class B : A
{
    public Rectangle getRectangle()
    {
        return [some other rectangle];
    }
}

The problem arises when I try something like this:

A a;
B b;
if(a.collision(b))
    ...

Where B's version of get rectangle is never actually called, as far as I can tell. I tried a solution like the one suggested here but the message I get is basically "B.getRectangle() hides inherited member A.getRectangle(). Use the new keyword if this was intended."

I appreciate in advance any help I receive. I'm thinking my past java experience is getting in the way of understanding how C# is different. I guess if anyone just knows of a good link that explains the differences between C# and java or just how C# works in this respect that could suffice.

Cheers.

Look at this - http://stackoverflow.com/questions/2744528/method-overloads-resolution-and-jon-skeets-brain-teasers — Kumar, Mar 03 '11 at 05:57

Ani · Answer 1 · 2011-03-03T05:45:39.443

6

Unlike Java, methods in C# are not marked virtual by default. What your current code is doing is hiding the getRectangle method: there is an implicit new modifier on the declaration of the method in the derived class.

You need to explicitly include the virtual modifier in the method-declaration in the base class:

public virtual Rectangle getRectangle() { ... }

and override it in the derived class with:

public override Rectangle getRectangle() { ... }

edited Mar 03 '11 at 05:45

answered Mar 03 '11 at 05:39

Ani

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Props for the implicit new. I didn't realize the language did that. – Ritch Melton Mar 03 '11 at 05:55
@Ritch: It's a conceptual thing. You don't have to think of it that way - it's just a way of avoiding the 'brittle base-class' problem. – Ani Mar 03 '11 at 06:36
I just assumed it wouldn't compile. I use resharper, so I've never tried either. – Ritch Melton Mar 03 '11 at 06:40

score 1 · Answer 2 · answered Mar 03 '11 at 05:38

1

You need to use the keyword override for the getRectangle method in class B like so:

public override Rectangle getRectangle()

answered Mar 03 '11 at 05:38

FreeAsInBeer

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Such a solution makes me feel silly. I'll give inheritance in C# a more in depth look before asking this kind of question again. Probably could have given the debugging a better try, too. Thanks. – DevynS Mar 03 '11 at 05:40
@DevynS: Your welcome. Please mark this as the answer and/or upvote if it helped you out. – FreeAsInBeer Mar 03 '11 at 05:42

score 0 · Answer 3 · answered Mar 03 '11 at 05:39

0

I don't know Java, but I assume they have some sort of virtual/override syntax for methods?

  public class A
  {
    virtual public Rectangle getRectangle()
    {
      return [some rectangle];
    }

    public bool collision(A other)
    {
      Rectangle rect1 = getRectangle();
      Rectangle rect2 = other.getRectangle();
      return rect1.Intersects(rect2);
    }
  }

  public class B : A
  {
     override public Rectangle getRectangle()
     { 
        return [some other rectangle];
     }
   }

answered Mar 03 '11 at 05:39

Ritch Melton

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"but I assume they have some sort of virtual/override syntax". No Java has implicit virtual and override. So all methods are virtual and method of same name always acts as override. – Euphoric Mar 03 '11 at 06:05
Oh wow. I figured they were somewhere between C++ virtual, and C#'s virtual, new, override, abstract overkill. – Ritch Melton Mar 03 '11 at 06:08

score 0 · Answer 4 · answered Mar 03 '11 at 05:39

0

I would suggest use Override to change the functionality of getRectangle() in B.

answered Mar 03 '11 at 05:39

sajoshi

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C# Calling an overriden method within the superclass

4 Answers4