c++ - 调用自动构造函数 : why is my type incomplete?

Question

在下面的代码中，我想使用默认构造函数{.data = value}，因为我希望我的类是POD。我不明白我在编译时收到的错误消息(llvm 或 gnu，c++11):

#include <type_traits>

class a {
  char data;
  static inline a create(char c) { return {.data = c}; } // this fails
  static inline a create2(char c) { a x; x.data = c; return x; } // this is OK
public:
  void init(char c) { *this = create(c); }
};

int main() {
  a s;
  s.init('x');
  return std::is_pod<a>::value;
}

有错误信息

t.cc:5:43: error: no matching constructor for initialization of 'a'
  static inline a create(char c) { return {.data = c}; }
                                          ^~~~~~~~~~~
t.cc:3:7: note: candidate constructor (the implicit copy constructor) not viable: cannot convert
      argument of incomplete type 'void' to 'const a &'

哪位好心人能给我解释一下，为什么我要用的时候a的类型不完整，为什么会被当作void处理？

Answer 1

您不能聚合初始化私有(private)成员。

来自 https://en.cppreference.com/w/cpp/language/aggregate_initialization

An aggregate is one of the following types: ... class type (typically, struct or union), that has no private or protected non-static data members

由于 a 是一个 class，而不是 struct，data 是 private.

将 data 声明为 public，或将类型声明为 struct 以默认为 public .

然后替换 static inline a create(char c) { return {.data = c}; }

with static inline a create(char c) { return a { c }; }

根据 https://en.cppreference.com/w/cpp/language/list_initialization

直接列表初始化(二)