C++: Why does a struct\class need a virtual method in order to be polymorphic?

Question

Following this question, I'm wondering why a struct\class in C++ has to have a virtual method in order to be polymorphic.

Forcing a virtual destructor makes sense, but if there's no destructor at all, why is it mandatory to have a virtual method?

Answer 1

Because the type of a polymorphic object in C++ is, basically, determined from the pointer to its vtable, which is the table of virtual functions. The vtable is, however, only created if there's at least one virtual method. Why? Because in C++, you never get what you didn't explicitly ask for. They call it "you don't have to pay for something you don't need". Don't need polymorphism? You just saved a vtable.

Answer 2

Forcing a virtual destructor makes sense

Exactly. To destruct a virtual class manually (via delete) through its base class you need a virtual destructor. (Now, as I’ve been reminded in the comments, this isn’t usually needed: rather than use manual memory management, one would rely on modern smart pointers which also work correctly with non-virtual destructors.)

So any class which acts as a polymorphic base class usually needs either a virtual destructor or virtual functions anyway.

And since having runtime polymorphism adds an overhead (the class needs to store an additional pointer to its virtual method table), the default is not to add it, unless necessary anyway: C++’ design philosophy is “you only pay for what you need”. Making every class have a virtual method table would run afoul of this principle.