Why std::shared_ptr calls destructors from base and derived classes, where delete calls only destructor from base class? [duplicate]

This question already has an answer here:

shared_ptr magic :) 3 answers

Why when using std::shared_ptr deallocation calls destructors from both base and derived classes when second example calls only destructor from base class?

class Base
{
public:
    ~Base()
    {
        std::cout << "Base destructor" << std::endl;
    }
};

class Derived : public Base
{
public:
    ~Derived()
    {
        std::cout << "Derived destructor" << std::endl;
    }
};

void virtual_destructor()
{
    {
        std::cout << "--------------------" << std::endl;
        std::shared_ptr<Base> sharedA(new Derived);
    }

    std::cout << "--------------------" << std::endl;
    Base * a = new Derived;
    delete a;
}

Output:

--------------------
Derived destructor
Base destructor
--------------------
Base destructor

I was expecting the same behaviour in both cases.

delete a is undefined behaviour, because the class Base does not have a virtual destructor and the "complete object" of *a (more accurately: the most-derived object containing *a) is not of type Base.

The shared pointer is created with a deduced deleter that deletes a Derived *, and thus everything is fine.

(The effect of the deduced deleter is to say delete static_cast<Derived*>(__the_pointer)).

If you wanted to reproduce the undefined behaviour with the shared pointer, you'd have to convert the pointer immediately:

// THIS IS AN ERROR
std::shared_ptr<Base> shared(static_cast<Base*>(new Derived));

In some sense, it is The Right Way for the shared pointer to behave: Since you are already paying the price of the virtual lookup for the type-erased deleter and allocator, it is only fair that you don't then also have to pay for another virtual lookup of the destructor. The type-erased deleter remembers the complete type and thus incurs no further overhead.

Matthieu M.

A missing piece to Kerrek SB's answer is how does the shared_ptr knows the type ?

The answer is that there are 3 types involved:

the static type of the pointer (shared_ptr<Base>)
the static type passed to the constructor
the actual dynamic type of the data

And shared_ptr does not know of the actual dynamic type, but knows which static type was passed to its constructor. It then practices type-erasure... but remembers somehow the type. An example implementation would be (without sharing):

template <typename T>
class simple_ptr_internal_interface {
public:
    virtual T* get() = 0;
    virtual void destruct() = 0;
}; // class simple_ptr_internal_interface

template <typename T, typename D>
class simple_ptr_internal: public simple_ptr_internal_interface {
public:
    simple_ptr_internal(T* p, D d): pointer(p), deleter(std::move(d)) {}

    virtual T* get() override { return pointer; }
    virtual void destruct() override { deleter(pointer); }

private:
    T* pointer;
    D deleter;
}; // class simple_ptr_internal

template <typename T>
class simple_ptr {
    template <typename U>
    struct DefaultDeleter {
        void operator()(T* t) { delete static_cast<U*>(t); }
    };

    template <typename Derived>
    using DefaultInternal = simple_ptr_internal<T, DefaultDeleter<Derived>>;

public:
    template <typename Derived>
    simple_ptr(Derived* d): internal(new DefaultInternal<Derived>{d}) {}

    ~simple_ptr() { this->destruct(); }

private:
    void destruct() { internal->destruct(); }

    simple_ptr_internal_interface* internal;
}; // class simple_ptr

Note that thanks to this mechanism, shared_ptr<void> is actually meaningful and can be used to carry any data an properly dispose of it.

Note also that there is a penalty involved with this semantics: the need for indirection required for the type-erasure of the deleter attribute.

来源：https://stackoverflow.com/questions/20802810/why-stdshared-ptr-calls-destructors-from-base-and-derived-classes-where-delet

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