c++ - 可变参数宏和参数包扩展

Question

在下面的示例代码中，我想使用带有可变参数的 MACRO_EXPANSION 和 {...} 来构造一个 EnumTypes 对象。但是，我无法使这个想法奏效。 (PS.代码结构可能看起来不太好，但我需要它:))

#include <iostream>
#include <utility>
#include <initializer_list>

enum class EnumOneTypes {
  One0,
  One1
};

enum class EnumTwoTypes {
  Two0,
  Two1
};

struct EnumTypes {
  EnumOneTypes one;
  EnumTwoTypes two;
};

void do_something(std::initializer_list<EnumTypes> il) {
    std::cout << "Do something" << std::endl;
}

// Need this struct to forward arguments
struct Register {
  template <typename... TArgs>
  Register(TArgs&&... args) {
    do_something(std::forward<TArgs>(args)...);
    //also do other things after do_something, omit here
    // ...
  }
};

// Use this macro to define global static objects
#define MACRO_EXPANSION(name, ...) \
  static struct Register name(__VA_ARGS__)

MACRO_EXPANSION(
  register_two,
  {EnumOneTypes::One0, EnumTwoTypes::Two0},
  {EnumOneTypes::One1, EnumTwoTypes::Two1}
);

MACRO_EXPANSION(
  register_three,
  {EnumOneTypes::One0, EnumTwoTypes::Two0},
  {EnumOneTypes::One1, EnumTwoTypes::Two1},
  {EnumOneTypes::One0, EnumTwoTypes::Two1}
);

int main() {
  std::cout << "Test the usage of this macro" << std::endl;
  return 0;
}

Answer 1

1. 可变参数模板不能自动成为 std::initializer_list。因此，让我们用大括号将可变参数括起来。

struct Register {
  template <typename... TArgs>
  Register(TArgs&&... args) {
    do_something({std::forward<TArgs>(args)...}); // Make params to be initializer List
    //also do other things after do_something, omit here
    // ...
  }
};

1. 由于 Register 构造函数是模板化的，因此编译器似乎无法推断出 {EnumOneTypes::One0, EnumTwoTypes::Two0} 属于哪种类型。所以让我们指定它的类型:

MACRO_EXPANSION(
  register_two,
  EnumTypes{EnumOneTypes::One0, EnumTwoTypes::Two0},
  EnumTypes{EnumOneTypes::One1, EnumTwoTypes::Two1}
);

应用这两个后，它编译成功并运行并输出:

Do something
Do something
Test the usage of this macro

我在 godbolt 中测试过.