Why you don't need virtual base destructor with smart pointers

struct Foo {
    // virtual ~Foo() {};
    int a;
};

struct Bar : public Foo {
    ~Bar() {std::cout << "bar dtor" << std::endl;};
};

int main() {
    std::shared_ptr<Foo> f = std::make_shared<Bar>();
    //Foo* f = new Bar();
    return 0;
}

In this example, the shared_ptr version would work as you expect. The raw pointer version will, however, not call Bar's destructor because Foo's dtor is not declared virtual. The question is why the shared_ptr version works?

Smart pointers work by managing a separate control block. It not only stores a ref count in the control block but also a deleter, a callable that gets called when ref count reaches zero. Multiple shared_ptrs of the same underlying resource must share the control block.

In the example above, the deleter of the control block is set when a shared_ptr<Bar> was created. Hence the deleter will always delete a Bar object, calling Bar's destructor, no matter what happens. It has nothing to do with virtual dispatch. Hence it doesn't require Foo to have a virtual destructor.

libc++'s implementation

Let's take a look at libc++'s implementation for the aforementioned behavior.

Notice how it always takes the pointer type and use that to create the deleter here https://github.com/llvm/llvm-project/blob/main/libcxx/include/__memory/shared_ptr.h#L670. This means the following would work as expected,

auto f = std::shared_ptr<Foo>(new Bar());

Tags: c++

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