SFINAE away a copy constructor
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).
The method of deriving from a copyable or non-copyable base is the standard idiom for this type of problem (see also Stefan's comment). One way to implement it is like this:
template<bool> struct copyable
{
protected:
~copyable() = default;
};
template<> struct copyable<false>
{
copyable(copyable const&) = delete;
copyable&operator=(copyable const&) = delete;
protected:
~copyable() = default;
};
template<bool allow_copy>
class some_class : copyable<allow_copy> { /* ... */ };