Can I list-initialize a vector of move-only type?

Solution 1:

Edit: Since @Johannes doesn't seem to want to post the best solution as an answer, I'll just do it.

#include <iterator>
#include <vector>
#include <memory>

int main(){
  using move_only = std::unique_ptr<int>;
  move_only init[] = { move_only(), move_only(), move_only() };
  std::vector<move_only> v{std::make_move_iterator(std::begin(init)),
      std::make_move_iterator(std::end(init))};
}

The iterators returned by std::make_move_iterator will move the pointed-to element when being dereferenced.

Original answer: We're gonna utilize a little helper type here:

#include <utility>
#include <type_traits>

template<class T>
struct rref_wrapper
{ // CAUTION - very volatile, use with care
  explicit rref_wrapper(T&& v)
    : _val(std::move(v)) {}

  explicit operator T() const{
    return T{ std::move(_val) };
  }

private:
  T&& _val;
};

// only usable on temporaries
template<class T>
typename std::enable_if<
  !std::is_lvalue_reference<T>::value,
  rref_wrapper<T>
>::type rref(T&& v){
  return rref_wrapper<T>(std::move(v));
}

// lvalue reference can go away
template<class T>
void rref(T&) = delete;

Sadly, the straight-forward code here won't work:

std::vector<move_only> v{ rref(move_only()), rref(move_only()), rref(move_only()) };

Since the standard, for whatever reason, doesn't define a converting copy constructor like this:

// in class initializer_list
template<class U>
initializer_list(initializer_list<U> const& other);

The initializer_list<rref_wrapper<move_only>> created by the brace-init-list ({...}) won't convert to the initializer_list<move_only> that the vector<move_only> takes. So we need a two-step initialization here:

std::initializer_list<rref_wrapper<move_only>> il{ rref(move_only()),
                                                   rref(move_only()),
                                                   rref(move_only()) };
std::vector<move_only> v(il.begin(), il.end());

Solution 2:

The synopsis of <initializer_list> in 18.9 makes it reasonably clear that elements of an initializer list are always passed via const-reference. Unfortunately, there does not appear to be any way of using move-semantic in initializer list elements in the current revision of the language.

Specifically, we have:

typedef const E& reference;
typedef const E& const_reference;

typedef const E* iterator;
typedef const E* const_iterator;

const E* begin() const noexcept; // first element
const E* end() const noexcept; // one past the last element

Solution 3:

As mentioned in other answers, the behaviour of std::initializer_list is to hold objects by value and not allow moving out, so this is not possible. Here is one possible workaround, using a function call where the initializers are given as variadic arguments:

#include <vector>
#include <memory>

struct Foo
{
    std::unique_ptr<int> u;
    int x;
    Foo(int x = 0): x(x) {}
};

template<typename V>        // recursion-ender
void multi_emplace(std::vector<V> &vec) {}

template<typename V, typename T1, typename... Types>
void multi_emplace(std::vector<V> &vec, T1&& t1, Types&&... args)
{
    vec.emplace_back( std::move(t1) );
    multi_emplace(vec, args...);
}

int main()
{
    std::vector<Foo> foos;
    multi_emplace(foos, 1, 2, 3, 4, 5);
    multi_emplace(foos, Foo{}, Foo{});
}

Unfortunately multi_emplace(foos, {}); fails as it cannot deduce the type for {}, so for objects to be default-constructed you have to repeat the class name. (or use vector::resize)

Solution 4:

Update for C++20: Using Johannes Schaub's trick of std::make_move_iterator() with C++20's std::to_array(), you can use a helper function like unto make_tuple() etc., here called make_vector():

#include <array>
#include <memory>
#include <vector>

struct X {};

template<class T, std::size_t N>
auto make_vector( std::array<T,N>&& a )
    -> std::vector<T>
{
    return { std::make_move_iterator(std::begin(a)), std::make_move_iterator(std::end(a)) };
}

template<class... T>
auto make_vector( T&& ... t )
{
    return make_vector( std::to_array({ std::forward<T>(t)... }) );
}

int main()
{
    using UX = std::unique_ptr<X>;
    const auto a  = std::to_array({ UX{}, UX{}, UX{} });     // Ok
    const auto v0 = make_vector( UX{}, UX{}, UX{} );         // Ok
    //const auto v2 = std::vector< UX >{ UX{}, UX{}, UX{} }; // !! Error !!
}

See it live on Godbolt.

Similar answer for older C++:

Using Johannes Schaub's trick of std::make_move_iterator() with std::experimental::make_array(), you can use a helper function:

#include <memory>
#include <type_traits>
#include <vector>
#include <experimental/array>

struct X {};

template<class T, std::size_t N>
auto make_vector( std::array<T,N>&& a )
    -> std::vector<T>
{
    return { std::make_move_iterator(std::begin(a)), std::make_move_iterator(std::end(a)) };
}

template<class... T>
auto make_vector( T&& ... t )
    -> std::vector<typename std::common_type<T...>::type>
{
    return make_vector( std::experimental::make_array( std::forward<T>(t)... ) );
}

int main()
{
    using UX = std::unique_ptr<X>;
    const auto a  = std::experimental::make_array( UX{}, UX{}, UX{} ); // Ok
    const auto v0 = make_vector( UX{}, UX{}, UX{} );                   // Ok
    //const auto v1 = std::vector< UX >{ UX{}, UX{}, UX{} };           // !! Error !!
}

See it live on Coliru.

Perhaps someone can leverage std::make_array()'s trickery to allow make_vector() to do its thing directly, but I did not see how (more accurately, I tried what I thought should work, failed, and moved on). In any case, the compiler should be able to inline the array to vector transformation, as Clang does with O2 on GodBolt.