How can I call const member function from destructor

Question

Is there any possible way to invoke const member function from destructor, when const object is destroyed?

Consider:

struct My_type {      ~My_type () {          show ();     }      void show () {          cout << "void show ()" << endl;     }     void show () const {          cout << "void show () const" << endl;     } }; 

And usage:

My_type mt; const My_type cmt; mt.show (); cmt.show (); 

Output:

void show () void show () const void show () void show () 

Can someone explain me why const version of show has not been invoked when cmt is destroyed?

Answer 1

The reason the non-const overload is being called when on a const instance is because cv-qualifiers on the current instance aren't considered during destruction. [class.dtor]/p2:

A destructor is used to destroy objects of its class type. The address of a destructor shall not be taken. A destructor can be invoked for a const, volatile or const volatile object. const and volatile semantics (7.1.6.1) are not applied on an object under destruction. They stop being in effect when the destructor for the most derived object (1.8) starts.

You can use a simply bind *this to a reference to const to get the behavior you need:

~My_type() {      My_type const& ref(*this);     ref.show(); } 

Or maybe you can use a wrapper that stores a reference and calls show() in its own destructor:

template<class T> struct MyTypeWrapper : std::remove_cv_t<T> {     using Base = std::remove_cv_t<T>;     using Base::Base;      T&       get()       { return ref; }     T const& get() const { return ref; }      ~MyTypeWrapper() { ref.show(); } private:     T& ref = static_cast<T&>(*this); };