Question

如何编写一个根据其参数类型执行不同操作的宏？

我有一个宏，该宏需要处理可以具有两种类型之一的参数。

#include <typeinfo>

enum class Coolness { kUndefined, kUncool, kCool };
enum class Tallness { kUndefined, kShort, kTall };

void MakePerson (Coolness coolness, Tallness tallness) {}

// Provide a way to make a person by only providing Coolness or Tallness.
#define MAKE_PERSON(x)                         \
({                                             \
  if (typeid(x) == typeid(Coolness)) {         \
      MakePerson(((x)), Tallness::kUndefined); \
  } else {                                     \
      MakePerson(Coolness::kUndefined, (x));   \
  }                                            \
})

int main()
{
  MAKE_PERSON(Coolness::kUncool);
  MAKE_PERSON(Tallness::kTall);
}

（我们可以在此处使用默认参数，但在实际代码中，我们实际上必须使用宏。）

编译器在main中的两个调用上均抛出错误：

main.cpp: In function ‘int main()’:
main.cpp:23:43: error: cannot convert ‘Coolness’ to ‘Tallness’ for argument ‘2’ to ‘void MakePerson(Coolness, Tallness)’
       MakePerson(Coolness::kUndefined, (x)); \
                                           ^
main.cpp:29:3: note: in expansion of macro ‘MAKE_PERSON’
   MAKE_PERSON(Coolness::kUncool);
   ^~~~~~~~~~~
main.cpp:21:45: error: cannot convert ‘Tallness’ to ‘Coolness’ for argument ‘1’ to ‘void MakePerson(Coolness, Tallness)’
       MakePerson(((x)), Tallness::kUndefined); \
                                             ^
main.cpp:30:3: note: in expansion of macro ‘MAKE_PERSON’
   MAKE_PERSON(Tallness::kTall);
   ^~~~~~~~~~~

（在https://www.onlinegdb.com/online_c++_compiler上完成）

我们不能像this question中那样使用__builtin_types_compatible_p，因为我们的编译器没有那个。

如何编写一个根据其参数类型执行不同操作的宏？

Answer 1

使用简单的函数重载，不要尝试使宏变得比需要的聪明：

enum class Coolness { kUndefined, kUncool, kCool };
enum class Tallness { kUndefined, kShort, kTall };

void MakePerson (Coolness coolness, Tallness tallness)
{
    ...
}

inline void MakePerson (Coolness coolness)
{
    MakePerson(coolness, Tallness::kUndefined);
}

inline void MakePerson (Tallness tallness)
{
    MakePerson(Coolness::kUndefined, tallness);
}

#define MAKE_PERSON(x) \
{ \
    // use __FILE__ and __LINE__ as needed... \
    MakePerson(x); \
}

int main()
{
    MAKE_PERSON(Coolness::kUncool);
    MAKE_PERSON(Tallness::kTall);
}

Live Demo

Answer 2

欢迎其他建议，但我们最终所做的是使用static_cast告诉编译器参数的类型：

#include <typeinfo>

enum class Coolness { kUndefined, kUncool, kCool };
enum class Tallness { kUndefined, kShort, kTall };

void MakePerson (Coolness coolness, Tallness tallness) {}

// Provide a way to make a person by only providing Coolness or Tallness.
// Static cast is used because the compiler fails to typecheck the 
// branches correctly without it.
#define MAKE_PERSON(x)                                              \
({                                                                  \
  if (typeid(x) == typeid(Coolness)) {                              \
      MakePerson(static_cast<Coolness>((x)), Tallness::kUndefined); \
  } else {                                                          \
      MakePerson(Coolness::kUndefined, static_cast<Tallness>((x))); \
  }                                                                 \
})

int main()
{
  MAKE_PERSON(Coolness::kUncool);
  MAKE_PERSON(Tallness::kTall);
}

...Program finished with exit code 0

如何使宏处理不同类型的参数？

2 个答案: