c++ - 为什么从初始化列表中初始化 vector 时不使用 move 构造(通过隐式构造函数)

Question

为了演示 move 语义，我编写了以下示例代码，其中包含来自 int 的隐式构造函数。

struct C {
  int i_=0;
  C() {}
  C(int i) : i_( i ) {}
  C( const C& other) :i_(other.i_) {
    std::cout << "A copy construction was made." << i_<<std::endl;
  }
  C& operator=( const C& other) {
    i_= other.i_ ;
    std::cout << "A copy assign was made."<< i_<<std::endl;
    return *this;
  }
  C( C&& other ) noexcept :i_( std::move(other.i_)) {
    std::cout << "A move construction was made." << i_ << std::endl;
  }
  C& operator=( C&& other ) noexcept {
    i_ = std::move(other.i_);
    std::cout << "A move assign was made." << i_ << std::endl;
    return *this;
  }
};

和

auto vec2 = std::vector<C>{1,2,3,4,5};
cout << "reversing\n";
std::reverse(vec2.begin(),vec2.end());

有输出

A copy construction was made.1
A copy construction was made.2
A copy construction was made.3
A copy construction was made.4
A copy construction was made.5
reversing
A move construction was made.1
A move assign was made.5
A move assign was made.1
A move construction was made.2
A move assign was made.4
A move assign was made.2

现在，反向显示了 2 次交换(每次使用一次 move 赋值和两次 move 构造)，但为什么从初始化列表创建的临时 C 对象不可能从中 move ？我以为我有一个整数的初始化列表，但我现在想知道我之间是否有一个 Cs 的初始化列表，不能从中 move (作为它的常量)。这是正确的解释吗？ - 发生了什么事？

Live demo

Answer 1

I thought I had an initializer list of integers, but I'm now wondering if what I have in between is an initializer list of Cs, which can't be moved from (as its const). Is this a correct interpretation?

这是正确的。 vector<C>没有 initializer_list<int>构造函数甚至 initializer_list<T>一些模板参数的构造函数 T .它确实有一个 initializer_list<C>构造函数 - 它由您传入的所有整数组成。自 initializer_list 的支持以来是一个 const 数组，你得到一堆拷贝而不是一堆 move 。