Question

Do interfaces (polymorphic class solely with pure virtual functions) have a vtable?
Since interfaces do not implement a polymorphic function themself and cant be directly constructed there would be no need for the linker to place a vtable. Is that so? Im especially concerned about the MSVC compiler.

Answer 1

Yes, they do. And there are a number of good reasons for that.

The first good reason is that even pure virtual methods have implementation. Either implicit or explicit. It is relatively easy to pull off a trick calling a pure virtual function, so you can basically provide a definition for one of yours, call it and see what happens. For that reason, there should be a virtual table in a first place.

There is another reason for putting a virtual table into a base class even if all of its methods are pure virtual and there are no other data members though. When polymorphism is used, a pointer to a base class is passed all around the program. In order to call a virtual method, compiler/runtime should figure out the relative offset of the virtual table from the base pointer. If C++ had no multiple inheritance, one could assume a zero offset from the abstract base class (for example), in which case it would have been possible not to have a vtable there (but we still need it due to reason #1). But since there is a multiple inheritance involved, a trick ala “vtable is there at 0 offset” won’t work because there could be two or three vtables depending on a number (and type) of base classes.

There could be other reasons I haven’t though of as well.

Hope it helps.