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如何计算可变参数__VA_ARGS__中的参数个数?
google上找了一下，网上答案不少，仔细看看原理都是一样的，比如这篇stackoverflow上的问题，给出的答案真不少，
《C preprocessor __VA_ARGS__ number of arguments》
但是排名第一的最佳答案有bug,在参数为空的时候计算错误。经测试找到正确计算方式，自己写了一个，代码如下，gcc上测试通过(MSV不支持)
代码语言：javascript复制// 计算 __VA_ARGS__ 参数个数,最大支持64个参数
#define FL_ARG_COUNT(...) FL_INTERNAL_ARG_COUNT_PRIVATE(0, ##__VA_ARGS__,
64, 63, 62, 61, 60,
59, 58, 57, 56, 55, 54, 53, 52, 51, 50,
49, 48, 47, 46, 45, 44, 43, 42, 41, 40,
39, 38, 37, 36, 35, 34, 33, 32, 31, 30,
29, 28, 27, 26, 25, 24, 23, 22, 21, 20,
19, 18, 17, 16, 15, 14, 13, 12, 11, 10,
9, 8, 7, 6, 5, 4, 3, 2, 1, 0)
#define FL_INTERNAL_ARG_COUNT_PRIVATE(
_0, _1, _2, _3, _4, _5, _6, _7, _8, _9,
_10, _11, _12, _13, _14, _15, _16, _17, _18, _19,
_20, _21, _22, _23, _24, _25, _26, _27, _28, _29,
_30, _31, _32, _33, _34, _35, _36, _37, _38, _39,
_40, _41, _42, _43, _44, _45, _46, _47, _48, _49,
_50, _51, _52, _53, _54, _55, _56, _57, _58, _59,
_60, _61, _62, _63, _64, N, ...) N
// 测试代码
FL_ARG_COUNT() ->0
FL_ARG_COUNT(1) ->1
FL_ARG_COUNT(1,2) ->2
FL_ARG_COUNT(1,2,3) ->3
FL_ARG_COUNT(1,2,3,4) ->4
FL_ARG_COUNT(1,2,3,4,5) ->5
FL_ARG_COUNT(1,2,3,4,5,6) ->6
FL_ARG_COUNT(1,2,3,4,5,6,7) ->7
FL_ARG_COUNT(1,2,3,4,5,6,7,8) ->8
FL_ARG_COUNT(1,2,3,4,5,6,7,8,9) ->9
FL_ARG_COUNT(1,2,3,4,5,6,7,8,9,10) ->10
FL_ARG_COUNT(
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4,5,6,7,8,9,10,
1,2,3,4
) ->64参考资料
《C preprocessor __VA_ARGS__ number of arguments》
《Macros with a Variable Number of Arguments.》
1:https://stackoverflow.com/questions/2124339/c-preprocessor-va-args-number-of-arguments
2:https://gcc.gnu.org/onlinedocs/gcc-5.2.0/gcc/Variadic-Macros.html#Variadic-Macros
