c++ - std::move 上发生不需要的转换

Question

为什么编译器认为我正在尝试将以下代码中Container的类型对象转换为类型对象？MoveTest为什么将大括号构造函数转发给move_things变量？替换{}为()当然可以解决问题

#include <iostream>
#include <array>
using namespace std;

static constexpr int NUMBER{2};

class MoveTest {
public:
  MoveTest(const MoveTest&) {
    cout << "MoveTest(const MoveTest&)" << endl;
  }
  MoveTest(MoveTest&&) {
    cout << "MoveTest(MoveTest&&)" << endl;
  }
  MoveTest() {
    cout << "MoveTest()" << endl;
  }
};

class Container {
public:
  Container() = default;
  Container(Container&&) = default;
  Container(const Container&) = default;
  std::array<MoveTest, NUMBER> move_things;
};

int main() {
  Container container;
  __attribute__((unused)) Container another_one{std::move(container)};

  return 0;
}

我得到的错误如下

Could not convert ‘std::move<Container&>((* & container))’ from ‘std::remove_reference<Container&>::type {aka Container}’ to ‘MoveTest`

这是否意味着统一初始化语法不再是任何带有std::array?

我使用的编译器和版本是

g++ (GCC) 4.8.5 20150623 (Red Hat 4.8.5-4)