std::tuple for non-copyable and non-movable object

问题

I have a class with copy & move ctor deleted.

struct A
{
    A(int a):data(a){}
    ~A(){ std::cout << "~A()" << this << " : " << data << std::endl; }

    A(A const &obj) = delete;
    A(A &&obj) = delete;

    friend std::ostream & operator << ( std::ostream & out , A const & obj);

    int data;
};

And I want to create a tuple with objects of this class. But the following does not compile:

auto p = std::tuple<A,A>(A{10},A{20}); 

On the other hand, the following does compile, but gives a surprising output.

int main() {
    auto q = std::tuple<A&&,A&&>(A{100},A{200});
    std::cout << "q created\n";
}

Output

~A()0x22fe10 : 100
~A()0x22fe30 : 200
q created

It means that dtor for objects is called as soon as tuple construction line ends. So, what is significance of a tuple of destroyed objects?

回答1:

This is bad:

auto q = std::tuple<A&&,A&&>(A{100},A{200});

you are constructing a tuple of rvalue references to temporaries that get destroyed at the end of the expression, so you're left with dangling references.

The correct statement would be:

std::tuple<A, A> q(100, 200);

---

However, until quite recently, the above was not supported by the standard. In N4296, the wording around the relevant constructor for tuple is [tuple.cnstr]:

template <class... UTypes>
  constexpr explicit tuple(UTypes&&... u);

Requires: sizeof...(Types) == sizeof...(UTypes). is_constructible<Ti, Ui&&>::value is true for all i.
Effects: Initializes the elements in the tuple with the corresponding value in std::forward<UTypes>(u).
Remark: This constructor shall not participate in overload resolution unless each type in UTypes is implicitly convertible to its corresponding type in Types.

So, this constructor was not participating in overload resolution because int is not implicitly convertible to A. This has been resolved by the adoption of Improving pair and tuple, which addressed precisely your use-case:

struct D { D(int); D(const D&) = delete; };    
std::tuple<D> td(12); // Error

The new wording for this constructor is, from N4527:

Remarks: This constructor shall not participate in overload resolution unless sizeof...(Types) >= 1 and is_constructible<Ti, Ui&&>::value is true for all i. The constructor is explicit if and only if is_convertible<Ui&&, Ti>::value is false for at least one i.

And is_constructible<A, int&&>::value is true.

To present the difference another way, here is an extremely stripped down tuple implementation:

struct D { D(int ) {} D(const D& ) = delete; };

template <typename T>
struct Tuple {
    Tuple(const T& t)
    : T(t)
    { }

    template <typename U,
#ifdef USE_OLD_RULES
              typename = std::enable_if_t<std::is_convertible<U, T>::value>
#else
              typename = std::enable_if_t<std::is_constructible<T, U&&>::value>
#endif
              >
    Tuple(U&& u)
    : t(std::forward<U>(u))
    { }

    T t;
};

int main()
{
    Tuple<D> t(12);
}

If USE_OLD_RULES is defined, the first constructor is the only viable constructor and hence the code will not compile since D is noncopyable. Otherwise, the second constructor is the best viable candidate and that one is well-formed.

---

The adoption was recent enough that neither gcc 5.2 nor clang 3.6 actually will compile this example yet. So you will either need a newer compiler than that (gcc 6.0 works) or come up with a different design.

回答2:

Your problem is that you explicitly asked for a tuple of rvalue references, and a rvalue reference is not that far from a pointer.

So auto q = std::tuple<A&&,A&&>(A{100},A{200}); creates two A objects, takes (rvalue) references to them, build the tuple with the references... and destroys the temporary objects, leaving you with two dangling references

Even if it is said to be more secure than good old C and its dangling pointers, C++ still allows programmer to write wrong programs.

Anyway, the following would make sense (note usage of A& and not A&&):

int main() {
    A a(100), b(100); // Ok, a and b will leave as long as main
    auto q = tuple<A&, A&>(a, b);  // ok, q contains references to a and b
    ...
    return 0; // Ok, q, a and b will be destroyed
}

来源：https://stackoverflow.com/questions/32763062/stdtuple-for-non-copyable-and-non-movable-object