Best style for deleting all constructors (or other function)?

问题

Let's say I want to make a type that cannot be constructed (don't ask why).

struct Impossible
{

I could do it like this:

    Impossible() = delete;
    // disable automatically generated constructors, don't declare any others
    Impossible(const Impossible&) = delete;
    // I suppose this is redundant with nothing to copy

or like this:

    Impossible(...) = delete;
    // explicitly disable all constructors

or like this:

    template<typename... Ts>
    Impossible(Ts...) = delete;
    // explicitly disable all constructors, template version
};

I guess I could ask the same thing about any function, not just constructors.

Does it make any difference which one I choose? In terms of syntax I think I like the second option. But is there any situation, whatsoever, where it's possible to detect a difference (other than in the text of an error message)?

回答1:

There is a difference, for example:

#include <type_traits>

struct Unconstructible
{
    Unconstructible() = delete;
};

static_assert( !std::is_default_constructible<Unconstructible>::value, "not default constructible" );
static_assert( std::is_copy_constructible<Unconstructible>::value, "copyable" );

Although you can never construct this object, so in practice you can never create a copy of one either, according the language and the type traits in the library, it is technically CopyConstructible, because there's an implicitly-declared copy constructor.

Similarly, with the Impossible(...) or Impossible(Ts&&...) forms there is still an implicitly-declared copy constructor.

On the other hand, if you do it this way:

#include <type_traits>

struct Unconstructible
{
    Unconstructible(const Unconstructible&) = delete;
};

static_assert( !std::is_default_constructible<Unconstructible>::value, "not default constructible" );
static_assert( !std::is_copy_constructible<Unconstructible>::value, "not copyable" );

The existence of a user-declared constructor suppresses the implicit declaration of the default constructor, so the class is neither DefaultConstructible nor CopyConstructible.

---

N.B. your last example should probably be Impossible(Ts&&...) to match any types, including non-copyable and non-movable ones.

回答2:

The first one is more than enough - no constructors will be generated by the compiler and, most imporatantly, it's idiomatic.

回答3:

You cannot create instance of this class:

struct Impossible
{
    Impossible() = delete;
};

Note the question of copy-constructor (or move-constructor) doesn't even arise, so deleting them is unnecessary. An object cannot exist because it cannot be constructed, so the copying or moving is out of question.

来源：https://stackoverflow.com/questions/14137853/best-style-for-deleting-all-constructors-or-other-function