Hide virtual method implementation if overloaded method is present in concrete implementation - c#

Here's the situation. I have a base class which implements a method that accepts a string, and sets a property to the value of that string. Under that, I have a group of concrete classes that implement the base. Some of the concrete classes can accept any string value to be stored. However, some of the class's values are constrained within an Enumerated series of values. The final value is still a string, but the value must be within that enumerated set of possible values.
Here's what I'd like to do: For the classes which use an enum-restricted value, I want to implement an overload to the string method that calls the string method with the correct value. However, I also want to lock the string implementation out so that only the enum implementation can be used to pass a new value in, in order to enforce the enum restriction.
I know at this point that this just isn't going to happen through the use of override or new in the method declaration. I also know that I could create a hierarchy like Base -> [BaseString|BaseEnum] -> Concrete and use sealing to keep the concrete implementation from allowing the string-based implementation to be used. What I'd like to know is if there's another way to do this that skips the extra level of hierarchy and allows me to just say "If method overload is present in concrete class, lock method."
public class Abstract {
public virtual string ValueHolder { get; private/protected set; }
public virtual void DoSomething( string s ) { ...; }
}
public class ConcreteEnum : Abstract {
public void DoSomething( Enum e ) {
base.DoSomething( "String somehow extracted from enum");
}
}
public class ConcreteString : Abstract { ... }
public class Main {
ConcreteString sc = new ConcreteString();
ConcreteEnum se = new ConcreteEnum();
sc.DoSomething( "I don't have an overload defined,
so I'm using my base's string handler method.
I can be any value, I have no restrictions." );
se.DoSomething( SomeEnum.MyValueWillBeRunThroughTheBaseStringHandler );
se.DoSomething( "I don't work because there's an
overload defined in my concrete class, and my
value is restricted to a specific set of
possibilities." );
}
As I said, I know that there are other viable ways of doing this, I just want to know if this is possible/more clean than those alternatives.

You would basically need to just validate the string in the derived class:
public class ConcreteEnum : Abstract {
public void DoSomething( Enum e ) {
base.DoSomething( "String somehow extracted from enum");
}
public override void DoSomething( string s ) {
// Validate s, then do something with it
}
}
You can't remove the method entirely, since it is a member of the superclass and hence part of the class interface. For example, if you passed an instance of ConcreteEnum to a method that takes an Abstract object and expected the DoSomething(string) method to work, but it was magically gone, all hell would break loose.
But you always have the option of overriding the behavior in a subclass to throw exceptions when an invalid value is supplied. You could even override it to just throw a NotSupportedException.
EDIT: If the base class DoSomething method is not supposed to be consumed except in subclasses (not saying it is, but you didn't really specify either way), why not just make it protected?

Related

C# Force method implementation in all sub-levels of inheritance and use base implementation

I have a BaseClass, which implements a method to populate itself form a different data structure. SubClasses will add their properties to the base ones. I want all sub-classes and sub-sub...classes to implement their own version of the method and call their parent class to do the same. So this is not just overriding, I want to force the implementation, but each implementation of the method has to be called, not just overridden.
Sort of like:
class BaseClass
{
int id;
virtual void fromDictionary(Dictionary data)
{
id = data["id"];
}
}
class Derived1 : BaseClass
{
string name;
override void fromDictionary(Dictionary data)
{
name = data["name"];
base.fromDictionary(data);
}
}
class Derived2 : Derived1
{
float size;
override void fromDictionary(Dictionary data)
{
size = data["size"];
base.fromDictionary(data);
}
}
Then doing this:
Derived2 object = new Derived2();
object.fromDictionary(dictionary);
Populates all of the object's properties.
I can make the first one virtual and override in the derived ones, but that doesn't force the implementation. And making an abstract base above this base or using an interface wouldn't force ALL levels of inheritance to implement.
The ideal case would be forcing all derived classes to implement their version AND call their parent's version (or even better, have some sort of extension/overriding that automatically calls ALL implementations of the method from the instantiated object's class upwards).
How close to that can I get ?
Seeing why you need this kind of overriding I strongly believe that you should try to move this logic to a constructor because as it looks now:
Derived2 object = new Derived2();
object.fromDictionary(dictionary);
Your object will only be valid if it has a dictionary. So instead of constructing it from a dictionary using a method, you should provide a constructor which receives a dictionary as a parameter:
Derived2 object = new Derived2(dictionary);
Now you have a valid object from the beggining. There are more reasons why you should do this instead of using a method which composes your object, one as you observed is that each subclass will need to call the base method, and having this kind of constructor (assuming that you will not provide a parameterless one) will force the inheritors to call the base one.
Another advantage in using this kind of approach is that you will have a valid object form the beginning instead of making it possible for users of that classes to make invalid objects by forgetting to call the fromDictionary() method.
Thanks for the suggestions everyone, the closest I could get is:
public abstract class DAO
{
public long id { get; set; }
public void fromDictionary(Dictionary<string, object> obj)
{
//Does own part in the method
id = (long)obj["id"];
//Calls most derived implementation
fromDictionaryOperation(obj);
}
//Forces child to implement its part
protected abstract void fromDictionaryOperation(Dictionary<string, object> obj);
}
//Is forced to implement its part, and the base implementation will be executed always
public class Area : DAO
{
public string name { get; set; }
protected override void fromDictionaryOperation(Dictionary<string, object> obj)
{
name = (string)obj["name"];
}
}
//Is NOT forced to implement method, and MUST call base.fromDictionary() for all this to work properly, but is NOT FORCED TO.
public class CircularArea : Area
{
public float radius { get; set; }
protected override void fromDictionaryOperation(Dictionary<string, object> obj)
{
radius = (float)obj["radius"];
base.fromDictionary(obj);
}
}
So all 2nd generation classes will be fine, but subsequent sub-classes wont be forced to implement its part or call the parent implementation. Which means that if in an implementation of a sub-sub-class, the base.fromDictionary() method is not called, then all parent classes implementation, except the first/base class, will be skipped without any compiling warning or error.
To force implementation at all levels, I guess I could put the abstract method in an Interface and make all classes implement the interface, which can't be forced itself, but is as close as I can think of.
If anyone knows a way to completely force ALL of them to implement the same method, that would be top notch.

How to define virtual method with return type which is not void in C#

This might sound like a dumb question, but I need to write a virtual method that is being overridden by inherited class. I don't need the virtual method to have any code, since this method is fully dependent on inherited class, therefore all code will be in the override methods.
However, the method has a return type that is not void. If I keep the virtual method empty it would give me an error "no all path return a value".
The only solution I came up with was to implement the virtual method with returning a dummy empty string, but I don't feel this is the best way. Is there any other way to define a virtual method with return type?
Edit:
Even most answers were correct in their own way, they did not help in my case, therefore I am adding snippets of the code which shows why I need to create instance of the base class, and why I can't use interface, or abstract:
//base class
public class Parser
{
public virtual string GetTitle()
{
return "";
}
}
//sub class
public class XYZSite : Parser
{
public override string GetTitle()
{
//do something
return title;
}
}
// in my code I am trying to create a dynamic object
Parser siteObj = new Parser();
string site = "xyz";
switch (site)
{
case "abc":
feedUrl = "www.abc.com/rss";
siteObj = new ABCSite();
break;
case "xyz":
feedUrl = "www.xzy.com/rss";
siteObj = new XYZSite();
break;
}
//further work with siteObj, this is why I wanted to initialize it with base class,
//therefore it won't break no matter what inherited class it was
siteObj.GetTitle();
I know the way I cast Parser object to Site object doesn't seem very optimal, but this is the only way it worked for me, so Please feel free to correct any thing you find wrong in my code.
Edit (Solution)
I followed the advice of many of replies by using interface and abstract. However it only worked for me when I changed the base class to abstract along with all its methods, and inherited the base class from the interface, and then inherited the sub classes from the base class. That way only I could make sure that all classes have the same methods, which can help me generate variant object in runtime.
Public interface IParser
{
string GetTitle();
}
Public abstract class Parser : IParser
{
public abstract string GetTitle();
}
Public class XYZ : Parser
{
public string GetTitle();
{
//actual get title code goes here
}
}
//in my web form I declare the object as follows
IParser siteObj = null;
...
//depending on a certain condition I cast the object to specific sub class
siteObj = new XYZ();
...
//only now I can use GetTitle method regardless of type of object
siteObj.GetTitle();
I am giving the credit to CarbineCoder since he was the one who put enough effort to take me the closest to the right solution. Yet I thank everyone for the contribution.
You can throw NotImplementedException instead of returning object:
public virtual object Method()
{
throw new NotImplementedException();
}
But if you are not implementing anything in virtual method you can create abstract instead of virtual:
public abstract object Method();
Edit:
Another option is to create interface for it.
public interface IMethods
{
object Method();
}
And make your classes children of this interface.
you need to use abstract here. The abstract modifier indicates that the thing being modified has a missing or incomplete implementation.
public abstract returntype MethodName();
But as you say, 'since this method is fully dependent on inherited class, therefore all code will be in the override methods', than if you are really going to override the functionality of the method in inherited class, why do you care if the method returns dummy or stuff? (e.g: you can make it virtual and get going)
Edit: as you cannot mark class as abstract, you can use virtual method instead.
public virtual returntype MethodName()
{
.....
return xyz;
}
(just for info: An abstract member is implicitly virtual. and abstract is sort of pure virtual. so you need virtual, instead of pure virtual)
Since other answers have discussed about abstract/virtual implementation, I am suggesting my own version.
There is a contradiction in your requirement.
You want a base class which is not an abstract but it has a method which is not implemented. Don't you think this unimplemented method will make the class incomplete and end up making it an abstract one even though you haven't explicitly said so?
So lets assume your class will never be an abstract class and its perfectly reasonable to have it as a normal class. Does it make sense to remove this method from the class altogether and move it to an interface?
Can you try extracting this method and put it into an interface.
interface NewInterface
{
string NewMethod();
}
public BaseClass
{
...
}
public DerivedClass : BaseClass, NewInterface
{
public string NewMethod
{
...
}
}
If you can do this, then you need not have to worry about the base class being abstract/ having NotImplemented exception, only downside is every derived class should implement this interface, but thats the point of making the base class non-abstract.
I don't see any problem in implementing Abstract BaseClass/ Interface for your approach. Both are supposed to be the solution for your problem.
//Parser siteObj = new Parser(); - Dont initialize it here,
//your are initializing it once more below
NewIterface siteObj;
string site = "xyz";
switch (site)
{
case "abc":
feedUrl = "www.abc.com/rss";
siteObj = new ABCSite();
break;
case "xyz":
feedUrl = "www.xzy.com/rss";
siteObj = new XYZSite();
break;
}

Base class object as argument for derived class

(Simplified) Scenario:
public class BaseClass
{
public int BaseClassInt {get; set;}
public BaseClass(int pBaseClassInt)
{ this.BaseClassInt = pBaseClassInt; }
}
public class DerivedClass : BaseClass
{
public int DerivedClassInt {get; set;}
public DerivedClass (int pBaseClassInt, int pDerivedClassInt) : base(pBaseClassInt)
{ this.DerivedClassInt = pDerivedClassInt; }
}
If I want to instantiate a DerivedClass-object I have to pass all arguments required to create a BaseClass-object and a DerivedClass-object. Also for every BaseClass-constructor I have to (at least should in my concrete case) provide a constructor with the same arguments in the derived class, plus arguments for the derived class properties. Then, if I change or delete a constructor in the base class I have to change or delete the corresponding contructor in the derived class(es).
I wonder if it is possible to use a constructor for the derived class which accepts a base class-object as an argument:
public DerivedClass(BaseClass pBaseClassObejct, int pDerivedClassInt)
{
// to make clear what I intend to do - looks silly of course
this = (DerivedClass)pBaseClassObject;
this.DerivedClassInt = pDerivedClassInt;
}
This could be called:
DerivedClass DerivedClassObject = new DerivedClass((new BaseClass(1),2);
If constructors in the base class would change, I wouldn´t have to mind it for the derived class. Is there any way to achieve this?
Think about this line for a moment:
this = (DerivedClass) pBaseClassObject;
Let's ignore the fact that you cant set this directly that way, and focus on the rest.
Imagine Giraffe and Elephant are both implementations of AfricanAnimal:
// By extension, ellie is also an AfricanAnimal
Elephant ellie = new Elephant();
// assume ellie is passed in as a param here (she can
// be, because she is an AfricanAnimal after all!):
public Giraffe(AfricanAnimal ellie)
{
this = (Giraffe) ellie; // Can't do this!
}
You can't (and would not want to) force ellie into being a giraffe, because a giraffe may have properties etc. that ellie lacks, and ellie may have properties that Giraffes don't have. Yet, using an AfricanAnimal as your parameter type there, would allow for just that.
Note: You could write that code and pass a Giraffe in, and all would be fine, but then again, that makes little sense; then you might as well use the Giraffe type as the parameter.
If you replace this with an instance variable, you would be able to compile with something like the following...
public Giraffe(AfricanAnimal ellie)
{
this.varOfTypeGiraffe = (Giraffe) ellie;
}
... but as soon as you run it with an Elephant as a a prameter, you will get an exception similar to:
InvalidCastException: Unable to cast object of type 'Elephant' to type 'Giraffe'.
TL;DR: This is a bad idea. Don't even try.
You cannot make a base constructor run from inside the body of any derived method (including the derived constructor). Even if you could, a base instance would not have retained any information about which constructor was used to instantiate it so there would be no way to know which base constructor should be called.
The above refers to the general case where a base constructor can potentially modify application state not directly related to the base class (e.g. by changing the value of static fields somewhere). You could use reflection to copy property values from a base instance to the derived instance being created, but this is practically unworkable because
It requires that you create a base instance in the first place -- what if the base is abstract, or if creating one has side effects?
You need a guarantee that the base constructor does not modify application state. But the aim here is to be independent of what the base constructors do, so you are back to square one.
No, that is not possible and should not be, because it doesn't make sense.
If it was possible and you deleted/changed the base class constructor, you would still need to change the code which creates the base class object that you would use as an argument to the derived class constructor.
Also, not all base classes are concrete. You would not be able to create an abstract base class, right?
This feature is not available. I think what you want is a little like this:
Suppose C# had a keyword allbaseargs and allowed code like this:
public class DerivedClass : BaseClass
{
public int DerivedClassInt { get; set; }
public DerivedClass (allbaseargs, int pDerivedClassInt)
: base(allbaseargs)
{
DerivedClassInt = pDerivedClassInt;
}
}
Then this could only work if BaseClass had only one (accessible) instance constructor.
The compiler should then examine the sole base constructor and substitute the magical word allbaseargs with the parameters of that constructor.
However, C# does not have this feature, and you would have to hand-code everything, which includes changeing all : base(...) calls of all derived classes when the constructor signature changes.
It is allowed to have the signature:
public DerivedClass(BaseClass pBaseClassObejct, int DerivedClassInt)
like you suggest, but you would not be able to chain the : base(...) easily. You would have to equip BaseClass with a construtor that took another instance in and copied all "state" (all instance properties and fields and such) from that other instance to "this". I do not recommend that solution.
This might be help!
Solution A: Create Inherit instead of base!
public static class Test
{
public static T Foo<T>(string text, int num) where T : BaseClass
{
T #base = (T)Activator.CreateInstance(typeof(T), new object[] { text, num });
//...
return #base;
}
public static void Main()
{
InheritClass inherit = Foo<InheritClass>("Hi there", 10);
}
}
Solution B: Copy base to inherit
public static class Test
{
public static TInherit As<TBase, TInherit>(this TBase #this) where TInherit : TBase
{
var type = typeof(TInherit);
var instance = Activator.CreateInstance(type);
foreach (var property in type.GetProperties())
if (property.CanWrite)
property.SetValue(instance, property.GetValue(#this, null), null);
return (TInherit)instance;
}
public static BaseClass Foo(string text, int num)
{
BaseClass #base = new BaseClass(text, num);
//...
return #base;
}
public static void Main()
{
InheritClass inherit = Foo("Hi there", 10).As<BaseClass, InheritClass>();
}
}
Notes: you can have simple 'As()' found here, but i prefer mine (where Inherit : TBase), where it's more safe and support converting base to inherit of inherit class.

Why would you mask a base class member?

I have just learned how to mask a base class member (using new) but am missing the point as to why I would want to do that. Does masking provide us with a certain level of protection as is the case in using encapsulation? Please advise.
You will very rarely use "new" to mask a base class member.
It's mainly used for the cases where the derived class had the member first, and then it was added to the base class --- the same name for a different purpose. The new is there to that you acknowledge that you know you are using it differently. When a base member is added in C++, it just silently merges the existing method into the inheritance chain. In C#, you will have to choose between new and override, to show you know what is happening.
It's not just used for masking. It actually breaks the inheritance chain, so if you call the base class method, the method in the derived class will not be called (just the one in the base class).
You're essentially creating a new method that has nothing to do with the base class method. Hence the "new" keyword.
Keeping that in mind the "new" keyword can be used if you want to define a method with the same signature as a base type method, but having a different return type.
The only valid safe examples that I've come across is being more specific with return types or providing a set accessor on a property. I'm not saying those are the only ones, but that's all I've found.
For example, suppose you have a very simple base that looks like this:
public abstract class Base
{
public string Name { get; protected set; }
public Base(string name)
{ Name = name; }
}
You could have a derived that looks more like this:
public class Derived : Base
{
public new string Name
{
get { return base.Name; }
set { base.Name = value; }
}
public Derived(string name) : base(name)
{ }
}
Assuming business rules allows this one specific Derived to have a changeable name, I believe this is acceptable. The problem with new is that it changes behavior depending on what type the instance is viewed as. For example, if I were to say:
Derived d = new Derived("Foo");
d.Name = "Bar";
Base b = d;
b.Name = "Baz"; // <-- No set available.
In this trivial example, we're fine. We are overriding the behavior with new, but not in a breaking way. Changing return types requires a bit more finesse. Namely, if you use new to change a return type on a derived type, you shouldn't allow that type to be set by the base. Check out this example:
public class Base
{
public Base(Base child)
{ Child = child; }
public Base Child { get; private set; }
}
public class Derived
{
public Derived(Derived child) : base(child)
{ }
public new Derived Child
{ get { return (Derived)base.Child; } }
}
If I could set Child on the Base class, I could have a casting problem in the Derived class. Another example:
Derived d = new Derived(someDerivedInstance);
Base b = d;
var c = b.Child; // c is of type Base
var e = d.Child; // e is of type Derived
I can't break any business rules by treating all of my Derived classes as Bases, it's just convenient to not type check and cast.
I have just learned how to mask a base class member (using new)
FYI this feature is usually called "hiding" rather than "masking". I think of "masking" as clearing bits in a bit array.
am missing the point as to why I would want to do that.
Normally you don't want to. For some reasons to use and not use this feature, see my article on the subject from 2008:
http://blogs.msdn.com/b/ericlippert/archive/2008/05/21/method-hiding-apologia.aspx
Does masking provide us with a certain level of protection as is the case in using encapsulation?
No, it does not.
What you are referring to is called Name Hiding. It is mostly a convenience feature. If you are inheriting from a class for which you do not control the source using new will let you change the behavior of a method even if it wasn't declared as virtual (or completely change the signature if it is virtual). The new keyword simply suppresses a compiler warning. You are basically informing the compiler that you are intentionally hiding the method from a parent class.
Delphi had the reintroduce keyword for the same reason.
What does this buy you other than a suppressed warning? Not a whole lot. You can't access the new method from a parent class. You can access it from an interface if your child class directly implements the interface (as apposed to inheriting it from its parent class). You can still call the parent class' member from the child. Any additional descendants of your class will inherit the new member rather than the one in the parent.
This is actually called member hiding. There are a couple of common scenarios where this can be appropriately used.
It allows you to work around versioning issues in which either the base or derived class author unwittingly creates a member name that collides with an existing identifier.
It can be used to simulate covariance on return types.
Regarding the first point...it is possible that an author of a base class could later add a member with the same name as an exisiting member in a derived class. The base class author may not have an knowledge of the derived classes and thus there is no expectation that she should avoid name collisions. C# supports the independent evolution of class hierarchies using the hiding mechanisms.
Regarding the second point...you may want a class to implement an interface that dictates a certain method signature and so you are locked into returning instances of a certain type only while at the same time you have subclassed that type and would really like for callers to see the concrete type instead. Consider this example.
public interface IFoo { }
public class ConcreteFoo { }
public abstract class Base
{
private IFoo m_Foo;
public Base(IFoo x) { m_Foo = x; }
public IFoo Foo { get { return m_Foo; } }
}
public class Derived
{
public Derived(ConcreteFoo x) : base(x) { }
public new ConcreteFoo Foo { get { return (ConcreteFoo)base.Foo; } }
}

Why does this work? Method overloading + method overriding + polymorphism

In the following code:
public abstract class MyClass
{
public abstract bool MyMethod(
Database database,
AssetDetails asset,
ref string errorMessage);
}
public sealed class MySubClass : MyClass
{
public override bool MyMethod(
Database database,
AssetDetails asset,
ref string errorMessage)
{
return MyMethod(database, asset, ref errorMessage);
}
public bool MyMethod(
Database database,
AssetBase asset,
ref string errorMessage)
{
// work is done here
}
}
where AssetDetails is a subclass of AssetBase.
Why does the first MyMethod call the second at runtime when passed an AssetDetails, rather than getting stuck in an infinite loop of recursion?
C# will resolve your call to your other implementation because calls to a method on an object, where the class for that object has its own implementation will be favored over an overridden or inherited one.
This can lead to subtle and hard-to-find problems, like you've shown here.
For instance, try this code (first read it, then compile and execute it), see if it does what you expect it to do.
using System;
namespace ConsoleApplication9
{
public class Base
{
public virtual void Test(String s)
{
Console.Out.WriteLine("Base.Test(String=" + s + ")");
}
}
public class Descendant : Base
{
public override void Test(String s)
{
Console.Out.WriteLine("Descendant.Test(String=" + s + ")");
}
public void Test(Object s)
{
Console.Out.WriteLine("Descendant.Test(Object=" + s + ")");
}
}
class Program
{
static void Main(string[] args)
{
Descendant d = new Descendant();
d.Test("Test");
Console.In.ReadLine();
}
}
}
Note that if you declare the type of the variable to be of type Base instead of Descendant, the call will go to the other method, try changing this line:
Descendant d = new Descendant();
to this, and re-run:
Base d = new Descendant();
So, how would you actually manage to call Descendant.Test(String) then?
My first attempt looks like this:
public void Test(Object s)
{
Console.Out.WriteLine("Descendant.Test(Object=" + s + ")");
Test((String)s);
}
This did me no good, and instead just called Test(Object) again and again for an eventual stack overflow.
But, the following works. Since, when we declare the d variable to be of the Base type, we end up calling the right virtual method, we can resort to that trickery as well:
public void Test(Object s)
{
Console.Out.WriteLine("Descendant.Test(Object=" + s + ")");
Base b = this;
b.Test((String)s);
}
This will print out:
Descendant.Test(Object=Test)
Descendant.Test(String=Test)
you can also do that from the outside:
Descendant d = new Descendant();
d.Test("Test");
Base b = d;
b.Test("Test");
Console.In.ReadLine();
will print out the same.
But first you need to be aware of the problem, which is another thing completely.
See the section of the C# Language Specification on Member Lookup and Overload Resolution. The override method of the derived class is not a candidate because of the rules on Member Lookup and the base class method is not the best match based on the Overload Resolution rules.
Section 7.3
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. If no members named N exist and are accessible, then the lookup produces no match, and the following steps are not evaluated.
Section 7.4.2:
Each of these contexts defines the set of candidate function members and the list of arguments in its own unique way, as described in detail in the sections listed above. For example, 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). (emphasis mine)
As others have correctly noted, when given the choice between two applicable candidate methods in a class, the compiler always chooses the one that was originally declared "closer" to the class which contains the call site when examining the base class hierarchy.
This seems counterintuitive. Surely if there is an exact match declared on a base class then this is a better match than an inexact match declared on a derived class, yes?
No. There are two reasons to choose the more derived method always over the less derived method.
The first is that the author of the derived class has much more information than the author of the base class. The author of the derived class knows everything about both the base class and the derived class, which is, after all, the class that the caller is actually using. When given the choice between calling a method written by someone who knows everything vs someone who knows only something about the type the caller is using, clearly it makes sense to prioritize calling the method written by the designer of the derived class.
Second, making that choice leads to a form of the Brittle Base Class Failure. We wish to protect you from this failure, and therefore have written the overload resolution rules so as to avoid it whenever possible.
For a detailed explanation of how this rule protects you from the Brittle Base Class Failure, see my article on the subject.
And for articles on other ways that languages deal with Brittle Base Class situations, click here.
Because that's the way the language is defined. For virtual members, the Implementation which is called at runtime, when a method exists in both a base class and a derived class, is based on the concrete type of the object which the method is called against, not the declared type of the variable which holds the reference to the object. Your first MyMethod is in an abstract class. So it can never be called from an object of type MyClass - because no such object can ever exist. All you can instanitate is derived class MySubClass. The concrete type is MySubClass, so that implementation is called, no matter that the code that calls it is in the base class.
For non-virtual members/methods, just the opposite is true.

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