What are the advantages of boost::noncopyable

Question

To prevent copying a class, you can very easily declare a private copy constructor / assignment operators. But you can also inherit boost::noncopyable.

Answer 1

I see no documentation benefit:

#include 

struct A
    : private boost::noncopyable
{
};

vs:

struct A
{
     A(const A&) = delete;
     A& operator=(const A&) = delete;
};

When you add move-only types, I even see the documentation as misleading. The following two examples are not copyable, though they are movable:

#include 

struct A
    : private boost::noncopyable
{
    A(A&&) = default;
    A& operator=(A&&) = default;
};

vs:

struct A
{
    A(A&&) = default;
    A& operator=(A&&) = default;
};

Under multiple inheritance, there can even be a space penalty:

#include 

struct A
    : private boost::noncopyable
{
};

struct B
    : public A
{
    B();
    B(const B&);
    B& operator=(const B&);
};

struct C
    : public A
{
};

struct D
    : public B,
      public C,
      private boost::noncopyable
{
};

#include 

int main()
{
    std::cout << sizeof(D) << '\n';
}

For me this prints out:

3

But this, which I believe to have superior documentation:

struct A
{
    A(const A&) = delete;
    A& operator=(const A&) = delete;
};

struct B
    : public A
{
    B();
    B(const B&);
    B& operator=(const B&);
};

struct C
    : public A
{
    C(const C&) = delete;
    C& operator=(const C&) = delete;
};

struct D
    : public B,
      public C
{
    D(const D&) = delete;
    D& operator=(const D&) = delete;
};

#include 

int main()
{
    std::cout << sizeof(D) << '\n';
}

Outputs:

2

I find it much easier to declare my copy operations than to reason whether or not I'm deriving from boost::non_copyable multiple times and if that is going to cost me. Especially if I'm not the author of the complete inheritance hierarchy.