Providing correct move semantics

Question

I am currently trying to figure out how to do move semantics correctly with an object which contains a pointer to allocated memory. I have a big datastructure, which contains an internal raw pointer to the actual storage (for efficiency reasons). Now I added a move constructor and move operator=(). In these methods I am std::move()ing the pointer to the new structure. However I am not sure what to do with the pointer from the other structure.

Here is a simple example of what I am doing:

class big_and_complicated {
   // lots of complicated code
};

class structure {
public:
   structure() :
      m_data( new big_and_complicated() )
   {}

   structure( structure && rhs ) :
      m_data( std::move( rhs.m_data ) )
   {
      // Maybe do something to rhs here?
   }

   ~structure()
   {
      delete m_data;
   }

private:
   big_and_complicated * m_data;
}

int main() {
  structure s1;
  structure s2( std::move( s1 ) );
  return 0;
}

Now from what I understand, after the std::move( s1 ) to s2 the only thing that is safe to do on s1 ist to call its constructor. However as far as I can see, this would lead to deleting the pointer contained within s1 in the destructor, rendering s2 useless as well. So I am guessing I have to do something to render the destructor safe when std::move()ing the pointer. As far as I can see the safest thing to do here, is to set it to 0 in the moved object, since this would turn the delete into a no-op later on. Is this reasoning correct so far? Or is std::move() actually smart enough to null out the pointer for me, rendering its usage safe? So far I am seeing no crashes in the actual test-suite, but I have not made sure the move-constructor is actually called.

Answer 1

"Moving" a pointer is no different than copying one and does not set the moved-from value to null ('moving' is in quotes here because std::move does not move anything really, it just changes the value category of the argument). Just copy rhs's pointer then set it to nullptr:

struct structure
{
    structure()
      : m_data{new big_and_complicated{}}
    { }

    structure(structure&& rhs)
      : m_data{rhs.m_data}
    {
        rhs.m_data = nullptr;
    }

    structure& operator =(structure&& rhs)
    {
        if (this != &rhs)
        {
            delete m_data;
            m_data = rhs.m_data;
            rhs.m_data = nullptr;
        }
        return *this;
    }

    ~structure()
    {
        delete m_data;
    }

private:
    big_and_complicated* m_data;

    structure(structure const&) = delete;             // for exposition only
    structure& operator =(structure const&) = delete; // for exposition only
}

You could also simplify this by using std::exchange:

structure(structure&& rhs)
    : m_data{ std::exchange(rhs.m_data, nullptr) }
{ }

structure& operator=(structure&& rhs)
{
    if (this != &rhs)
    {
        delete m_data;
        m_data = std::exchange(rhs.m_data, nullptr);
    }
    return *this;
}

Better yet, use std::unique_ptr<big_and_complicated> instead of big_and_complicated* and you don't need to define any of this yourself:

#include <memory>

struct structure
{
    structure()
      : m_data{new big_and_complicated{}}
    { }

private:
    std::unique_ptr<big_and_complicated> m_data;
}

Lastly, unless you actually want structure to remain non-copyable, you're better off just implementing proper move semantics inside of big_and_complicated and having structure hold a big_and_complicated object directly.