Why are destructors not virtual by default [C++]

Question

Why doesn't C++ make destructors virtual by default for classes that have at least one other virtual function? In this case adding a virtual destructor costs me nothing, and not having one is (almost?) always a bug. Will C++0x address this?

Answer 1

You don't pay for what you don't need. If you never delete through base pointer, you may not want the overhead of the indirected destructor call.

Perhaps you were thinking that the mere existence of the vtable is the only overhead. But each individual function dispatch has to be considered, too, and if I want to make my destructor call dispatch directly, I should be allowed to do so.

It would be nice of your compiler to warn you if you do ever delete a base pointer and that class has virtual methods, I suppose.

Edit: Let me pull Simon's excellent comment in here: Check out this SO question on the code generated for destructors. As you can see, there's also code-bloat overhead to be considered.

Answer 2

Here's an example (not that I recommend writing such code):

struct base {     virtual void foo() const = 0;     virtual void bar () const = 0; };  struct derived: base {     void foo() const {}     void bar() const {} };  std::shared_ptr<base> make_base() {     return std::make_shared<derived>(); } 

This is perfectly fine code that does not exhibit UB. This is possible because std::shared_ptr uses type-erasure; the final call to delete will delete a derived*, even if the last std::shared_ptr to trigger destruction is of type std::shared_ptr<void>.

Note that this behaviour of std::shared_ptr is not tailored to virtual destruction; it has a variety of other uses (e.g. std::shared_ptr<FILE> { std::fopen( ... ), std::fclose }). However since this technique already pays the cost of some indirection to work, some users may not be interested in having a virtual destructor for their base classes. That's what "pay only for what you need" means.