This is a pure virtual function. This means, that subclasses have to implement this function, otherwise they are abstract, meaning you cannot create objects of that class. class ISomeInterface { public: virtual std::string ToString( ) = 0; } class SomeInterfaceImpl : public ISomeInterface { public: virtual std::string ToString( ) { return “SomeInterfaceImpl”; } } The …
So basically if in your ancestor class you want a certain behaviour for a method. If your descendent uses the same method but has a different implementation you can override it, If it has a virtual keyword. using System; class TestClass { public class Dimensions { public const double pi = Math.PI; protected double x, …
You do get the surprising result that if you have a child, you can’t call foo, but you can cast it to a base and then call foo. child *c = new child(); c->foo; // compile error (can’t access private member) static_cast<base *>(c)->foo(); // this is fine, but still calls the implementation in child I …
No, because it doesn’t make any sense in C++. Virtual functions are invoked when you have a pointer/reference to an instance of a class. Static functions aren’t tied to a particular instance, they’re tied to a class. C++ doesn’t have pointers-to-class, so there is no scenario in which you could invoke a static function virtually.
Yes, you can definitely use = default for such destructors. Especially if you were just going to replace it with {}. I think the = default one is nicer, because it’s more explicit, so it immediately catches the eye and leaves no room for doubt. However, here are a couple of notes to take into …
Typically final will not be used on the base class’ definition of a virtual function. final will be used by a derived class that overrides the function in order to prevent further derived types from further overriding the function. Since the overriding function must be virtual normally it would mean that anyone could override that …
final does not require the function to override anything in the first place. Its effect is defined in [class.virtual]/4 as If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from B a function D::f overrides B::f, the program is ill-formed. That’s it. Now …
In C# you have to mark method as virtual to make it possible to override. Does it mean that in C# you should mark all methods virtual (except a few ones that you don’t want to be overridden), since most likely you don’t know in what way your class can be inherited? No. If the …
The assignment operator is not required to be made virtual. The discussion below is about operator=, but it also applies to any operator overloading that takes in the type in question, and any function that takes in the type in question. The below discussion shows that the virtual keyword does not know about a parameter’s …
In some cases, yes, it is legal for a derived class to override a virtual function using a different return type as long as the return type is covariant with the original return type. For example, consider the following: class Base { public: virtual ~Base() {} virtual Base* clone() const = 0; }; class Derived: …