聚合初始化原子成员

Question

似乎无法使用 C++14 初始化聚合中的原子成员。以下不起作用 (live on gcc 8.0.1):

#include <atomic>
#include <iostream>

struct stru {
  std::atomic_int32_t val_0;
  std::atomic_int32_t val_1;
};

int main() {
  auto p = new stru{0, 1};
  std::cout << p->val_0 << ", " << p->val_1 << std::endl; 
}

错误信息：

error: use of deleted function 'std::atomic<int>::atomic(const std::atomic<int>&)'
   auto p = new stru{0, 1};
                     ^

这是因为原子类型既不可复制也不可移动，因此不可复制初始化。然而，以下似乎有效（live on gcc 8.0.1）。

#include <atomic>
#include <iostream>

struct stru {
  std::atomic_int32_t val_0;
  std::atomic_int32_t val_1;
};

int main() {
  auto p = new stru{};
  std::cout << p->val_0 << ", " << p->val_1 << std::endl; 
}

这有效地执行了零初始化，因此无法初始化为零以外的值。有没有办法初始化为其他指定值？

Answer 1

1 和 0 不是 std::atomic_int32_t。以下作品：

#include <atomic>
#include <iostream>

struct stru {
  std::atomic_int32_t val_0;
  std::atomic_int32_t val_1;
};

int main() {
  auto p = new stru{{0}, {1}};
  std::cout << p->val_0 << ", " << p->val_1 << std::endl; 
}

编辑：为什么会这样？ Copy initialization kicks in 对于初始化列表中的每个元素：

Each direct public...non-static class member, in order of array subscript/appearance in the class definition, is copy-initialized from the corresponding clause of the initializer list.

在您的示例中，1 和 0 被隐式转换为 std::atomic_int32_t，并且 std::atomic_int32_t 不能从 [=] 变为 copy-initialized 13=]。在额外的 {} 版本中，val_0 和 val_1 是从 std::initializer_list<int> 复制初始化的，这很好。

Answer 2

Patrick 的 works, but his explanation doesn't look right to me. So I post the explanation I find here. For the code auto p = new stru{{0}, {1}}; aggregate initialization 效果：

If the initializer clause is a nested braced-init-list (which is not an expression), the corresponding array element/class member/public base (since C++17) is list-initialized from that clause: aggregate initialization is recursive.

因此，成员不是复制初始化，而是从 braced-init-list 复制 list-initialized。

聚合初始化原子成员

Initialize atomic members in aggregate

c++

c++14

language-lawyer

stdatomic

aggregate-initialization