SFINAE away a copy constructor

Question

Under certain conditions, I'd like to SFINAE away the copy constructor and copy assignment operator of a class template. But if I do so, a default copy constructor and a default assignment operator are generated. The SFINAE is done based on tags I pass as class template parameters. The problem is, that SFINAE only works on templates and a copy constructor/assignment operator can't be a template. Does there exist a workaround?

Answer 1

This solution uses a base class that is conditionally not copyable (by explicitely marking the copy constructor and copy assignment operator as deleted).

template <bool>
struct NoCopy;

template <>
struct NoCopy<true>
{
   // C++11 and later: marking as deleted. Pre-C++11, make the copy stuff private.
   NoCopy(const NoCopy&) = delete;
   NoCopy& operator=(const NoCopy&) = delete;
   protected:
      ~NoCopy() = default; // prevent delete from pointer-to-parent
};

template <>
struct NoCopy<false>
{
   // Copies allowed in this case
   protected:
      ~NoCopy() = default; // prevent delete from pointer-to-parent
};

Example usage:

template <typename Number>
struct Foo : NoCopy<std::is_integral<Number>::value>
{
   Foo() : NoCopy<std::is_integral<Number>::value>{}
   {
   }
};

int main()
{
   Foo<double> a;
   auto b = a; // compiles fine
   Foo<int> f;
   auto g = f; // fails!
}

Note: the destructor of NoCopy is declared protected to avoid virtual inheritance (Thanks for the hint, @Yakk).