我可以列出初始化只移动类型的向量吗?
如果我通过我的 GCC 4.7 快照传递以下代码，它会尝试将 unique_ptrs 复制到向量中.
If I pass the following code through my GCC 4.7 snapshot, it tries to copy the unique_ptrs into the vector.
#include <vector>
#include <memory>
int main() {
using move_only = std::unique_ptr<int>;
std::vector<move_only> v { move_only(), move_only(), move_only() };
}
显然这不起作用，因为 std::unique_ptr 不可复制:
Obviously that cannot work because std::unique_ptr is not copyable:
错误:使用已删除的函数 'std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = int;_Dp = std::default_delete;std::unique_ptr<_Tp, _Dp> = std::unique_ptr]'
error: use of deleted function 'std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = int; _Dp = std::default_delete; std::unique_ptr<_Tp, _Dp> = std::unique_ptr]'
GCC 在尝试从初始化列表中复制指针时是否正确?
Is GCC correct in trying to copy the pointers from the initializer list?
推荐答案
18.9 中 <initializer_list> 的概要清楚地表明，初始化列表的元素总是通过 const-reference 传递.不幸的是，在当前版本的语言中，似乎没有任何方法可以在初始化列表元素中使用移动语义.
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.
具体来说，我们有:
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
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