c++ - C++的这些宏怎么理解？

Question

C++的这些宏怎么理解
下面代码出自：https://github.com/sharkdp/dbg-macro的dbg.h头文件中

#define DBG_IDENTITY(x) x
#define DBG_CALL(fn, args) DBG_IDENTITY(fn args)

#define DBG_CAT_IMPL(_1, _2) _1##_2
#define DBG_CAT(_1, _2) DBG_CAT_IMPL(_1, _2)

#define DBG_16TH_IMPL(_1, _2, _3, _4, _5, _6, _7, _8, _9, _10, _11, _12, _13, \
                      _14, _15, _16, ...)                                     \
  _16
#define DBG_16TH(args) DBG_CALL(DBG_16TH_IMPL, args)
#define DBG_NARG(...) \
  DBG_16TH((__VA_ARGS__, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0))

// DBG_VARIADIC_CALL(fn, data, e1, e2, ...) => fn_N(data, (e1, e2, ...))
#define DBG_VARIADIC_CALL(fn, data, ...) \
  DBG_CAT(fn##_, DBG_NARG(__VA_ARGS__))(data, (__VA_ARGS__))

// (e1, e2, e3, ...) => e1
#define DBG_HEAD_IMPL(_1, ...) _1
#define DBG_HEAD(args) DBG_CALL(DBG_HEAD_IMPL, args)

// (e1, e2, e3, ...) => (e2, e3, ...)
#define DBG_TAIL_IMPL(_1, ...) (__VA_ARGS__)
#define DBG_TAIL(args) DBG_CALL(DBG_TAIL_IMPL, args)

#define DBG_MAP_1(fn, args) DBG_CALL(fn, args)
#define DBG_MAP_2(fn, args) fn(DBG_HEAD(args)), DBG_MAP_1(fn, DBG_TAIL(args))
#define DBG_MAP_3(fn, args) fn(DBG_HEAD(args)), DBG_MAP_2(fn, DBG_TAIL(args))
#define DBG_MAP_4(fn, args) fn(DBG_HEAD(args)), DBG_MAP_3(fn, DBG_TAIL(args))
#define DBG_MAP_5(fn, args) fn(DBG_HEAD(args)), DBG_MAP_4(fn, DBG_TAIL(args))
#define DBG_MAP_6(fn, args) fn(DBG_HEAD(args)), DBG_MAP_5(fn, DBG_TAIL(args))
#define DBG_MAP_7(fn, args) fn(DBG_HEAD(args)), DBG_MAP_6(fn, DBG_TAIL(args))
#define DBG_MAP_8(fn, args) fn(DBG_HEAD(args)), DBG_MAP_7(fn, DBG_TAIL(args))
#define DBG_MAP_9(fn, args) fn(DBG_HEAD(args)), DBG_MAP_8(fn, DBG_TAIL(args))
#define DBG_MAP_10(fn, args) fn(DBG_HEAD(args)), DBG_MAP_9(fn, DBG_TAIL(args))
#define DBG_MAP_11(fn, args) fn(DBG_HEAD(args)), DBG_MAP_10(fn, DBG_TAIL(args))
#define DBG_MAP_12(fn, args) fn(DBG_HEAD(args)), DBG_MAP_11(fn, DBG_TAIL(args))
#define DBG_MAP_13(fn, args) fn(DBG_HEAD(args)), DBG_MAP_12(fn, DBG_TAIL(args))
#define DBG_MAP_14(fn, args) fn(DBG_HEAD(args)), DBG_MAP_13(fn, DBG_TAIL(args))
#define DBG_MAP_15(fn, args) fn(DBG_HEAD(args)), DBG_MAP_14(fn, DBG_TAIL(args))
#define DBG_MAP_16(fn, args) fn(DBG_HEAD(args)), DBG_MAP_15(fn, DBG_TAIL(args))

// DBG_MAP(fn, e1, e2, e3, ...) => fn(e1), fn(e2), fn(e3), ...
#define DBG_MAP(fn, ...) DBG_VARIADIC_CALL(DBG_MAP, fn, __VA_ARGS__)

#define DBG_STRINGIFY_IMPL(x) #x
#define DBG_STRINGIFY(x) DBG_STRINGIFY_IMPL(x)

我尝试用渐进的方式去理解，尝试如下代码去理解前两行宏

#define DBG_IDENTITY(x) x
#define DBG_CALL(fn, args) DBG_IDENTITY(fn args)

void test(int x)
{
    std::cout << x << std::endl;
}
int main()
{
    DBG_CALL(test,2);
}

发现代码不可编译，故尝试问之

Answer 1

先说结论：改成 DBG_CALL(test, (2)); 可以通过了。

宏替换过程：

DBG_CALL(test, (2));
    --> DBG_IDENTITY(test (2));
        --> test(2);

问题来了，为什么不直接写#define DBG_CALL(fn, args) fn(args)? 因为某些情况微软的编译器无法正常处理。比如 arg 是 1,2,3 这种形式，具体可参考 github 代码注释里的连接。