Deleter – std::shared_ptr vs. std::unique_ptr

For the std::unique_ptr the deleter is part of the type and the std::shared_ptr keeps it as part of its control block. Therefore in a Base-Derived-Class szenario, with a non virtual destructor in the base class, the std::unique_ptr will just delete the base class part and the std::shared_ptr also the derived class part.

#include <iostream>
#include <memory>

struct Base
{
  ~Base()
  {
    std::cout << __FUNCTION__ << std::endl;
  }
};

struct Derived : public Base
{
  ~Derived()
  {
    std::cout << __FUNCTION__ << std::endl;
  }
};

int main()
{
  {
    std::cout << "unique_ptr" << std::endl;
    std::unique_ptr<Base> uPtr(new Derived);
  }
  std::cout << std::endl;
  {
    std::cout << "shared_ptr" << std::endl;
    std::shared_ptr<Base> sPtr(new Derived);
  }
  return 0;
}

$ g++ main.cpp -std=c++11
$ ./a.out
unique_ptr
~Base

shared_ptr
~Derived
~Base

Never the less please always make the destructor in the base class virtual.