I can be argued that the Most important distinction about virtual
, is that this causes the compiler, when calling a method member polymorphically, which implementation to bind the compiled code to. When you call a member of a class from client code where the actual type of the object is, say derived class foo
, but the variable it is being called on is actually typed (declared) as some base class, say bar
, Members declared as virtual
will bind to the implementation in the actual object type, (or to the most derived base class of the objects type that has an implementation). Members not declared as virtual will bind to the implementation in the type that the variable is declared to be.
A. Virtual. then, if the member is declared as virtual, the implementation in the derived class will be executed even if the variable is declared as a base type.
public class Animal
{ public virtual Move() { debug.Print("Animal.Move()"); }
public class Bird: Animal
{ public virtual override Move() { debug.Print("Bird.Move()"); }
Animal x = new Bird();
x.Move(); // Will print "Bird.Move"
B. Not Virtual. When a member which is not declared as virtual
, then the implementation will be chosen based on the declared type of the variable the method is executed on. So if you have a Bird Object, in variable x declared as `Animal', and you call a method that is implemented in both classes, the compiler will bind to the implementation in the Animal class, not in Bird, even though the object is really a Bird.
public class Animal
{ public Move() { debug.Print("Animal.Move()"); }
public class Bird: Animal
{ public Move() { debug.Print("Bird.Move()"); }
Animal x = new Bird();
x.Move(); // Will print "Animal.Move"