Macro to count number of arguments

#define PP_ARG_N( \ _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, N, ...) N #define PP_RSEQ_N() \ 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 PP_NARG_(...) PP_ARG_N(__VA_ARGS__) #define PP_NARG(...) PP_NARG_(__VA_ARGS__, PP_RSEQ_N()) #define PP_COMMASEQ_N() \ 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, \ 1, 1, 1, 1, 1, 1, 1, 1, 0, 0 #define PP_COMMA() , #define PP_HASCOMMA(...) \ PP_NARG_(__VA_ARGS__, PP_COMMASEQ_N()) #define PP_NARG(...) \ PP_NARG_HELPER1( \ PP_HASCOMMA(__VA_ARGS__), \ PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()), \ PP_NARG_(__VA_ARGS__, PP_RSEQ_N())) #define PP_NARG_HELPER1(a, b, N) PP_NARG_HELPER2(a, b, N) #define PP_NARG_HELPER2(a, b, N) PP_NARG_HELPER3_ ## a ## b(N) #define PP_NARG_HELPER3_01(N) 0 #define PP_NARG_HELPER3_00(N) 1 #define PP_NARG_HELPER3_11(N) N PP_NARG() // expands to 0 PP_NARG(x) // expands to 1 PP_NARG(x, 2) // expands to 2

Explanation:

The trick in these macros is that PP_HASCOMMA(...) expands to 0 when called with zero or one argument and to 1 when called with at least two arguments. To distinguish between these two cases, I used PP_COMMA __VA_ARGS__ (), which returns a comma when __VA_ARGS__ is empty and returns nothing when __VA_ARGS__is non-empty.

Now there are three possible cases:

__VA_ARGS__ is empty: PP_HASCOMMA(__VA_ARGS__) returns 0 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ()) returns 1.

__VA_ARGS__ contains one argument: PP_HASCOMMA(__VA_ARGS__) returns 0 andPP_HASCOMMA(PP_COMMA __VA_ARGS__ ()) returns 0.

__VA_ARGS__ contains two or more arguments: PP_HASCOMMA(__VA_ARGS__) returns 1 and PP_HASCOMMA(PP_COMMA __VA_ARGS__ ()) returns 1.

The PP_NARG_HELPERx macros are just needed to resolve these cases.

In order to fix the func(0, ) problem, we need to test whether we have supplied zero or more arguments. The PP_ISZERO macro comes into play here.

#define PP_ISZERO(x) PP_HASCOMMA(PP_ISZERO_HELPER_ ## x) #define PP_ISZERO_HELPER_0 ,

Now let's define another macro which prepends the number of arguments to an argument list:

#define PP_PREPEND_NARG(...) \ PP_PREPEND_NARG_HELPER1(PP_NARG(__VA_ARGS__), __VA_ARGS__) #define PP_PREPEND_NARG_HELPER1(N, ...) \ PP_PREPEND_NARG_HELPER2(PP_ISZERO(N), N, __VA_ARGS__) #define PP_PREPEND_NARG_HELPER2(z, N, ...) \ PP_PREPEND_NARG_HELPER3(z, N, __VA_ARGS__) #define PP_PREPEND_NARG_HELPER3(z, N, ...) \ PP_PREPEND_NARG_HELPER4_ ## z (N, __VA_ARGS__) #define PP_PREPEND_NARG_HELPER4_1(N, ...) 0 #define PP_PREPEND_NARG_HELPER4_0(N, ...) N, __VA_ARGS__

The many helpers are again needed to expand the macros to numeric values. Finally test it:

#define my_func(...) func(PP_PREPEND_NARG(__VA_ARGS__)) my_func() // expands to func(0) my_func(x) // expands to func(1, x) my_func(x, y) // expands to func(2, x, y) my_func(x, y, z) // expands to func(3, x, y, z)

It is possible to do in GCC using the ##VA_ARGS extension:

#define PP_ARG_N( \ _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, N, ...) N /* Note 63 is removed */ #define PP_RSEQ_N() \ 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 PP_NARG_(...) PP_ARG_N(__VA_ARGS__) /* Note dummy first argument _ and ##__VA_ARGS__ instead of __VA_ARGS__ */ #define PP_NARG(...) PP_NARG_(_, ##__VA_ARGS__, PP_RSEQ_N()) #define my_func(...) func(PP_NARG(__VA_ARGS__), __VA_ARGS__)

Now PP_NARG(a, b, c) gives 3 and PP_NARG() gives 0.

Unfortunately I don't see a way to make it work in general.

I came up with the following workaround for PP_NARG:

#define PP_NARG(...) (sizeof(#__VA_ARGS__) - 1 ? \ PP_NARG_(__VA_ARGS__, PP_RSEQ_N()) : 0)

It stringifies __VA_ARGS__, so if it's empty its length equals 1 (because#__VA_ARGS__ == '\0').It works with -std=c99 -pedantic.

I still have problems with wrapping the variadic function though. When__VA_ARGS__ is empty, my_func expands to func(0, ) which triggers a compilation error.

I usually use this macro to find a number of params:

#define NUMARGS(...) (sizeof((int[]){__VA_ARGS__})/sizeof(int))

So in case of GCC you need to define macros like this:

#define NUMARGS(...) (sizeof((int[]){0, ##__VA_ARGS__})/sizeof(int)-1) #define SUM(...) sum(NUMARGS(__VA_ARGS__), ##__VA_ARGS__)

counting args with C macros

/** * we need a placeholder at the start for zero so ##_VA_ARGS__ has something to consume * then the arguments are pushed out in such a way that 'cnt' ends up with the right count. */ #define COUNT_ARGS(...) COUNT_ARGS_(Z,##__VA_ARGS__,6,5,4,3,2,1,0) #define COUNT_ARGS_(z,a,b,c,d,e,f,cnt,...) cnt #define C_ASSERT(test) \ switch(0) {\ case 0:\ case test:;\ } int main() { C_ASSERT(0 == COUNT_ARGS()); C_ASSERT(1 == COUNT_ARGS(a)); C_ASSERT(2 == COUNT_ARGS(a,b)); C_ASSERT(3 == COUNT_ARGS(a,b,c)); C_ASSERT(4 == COUNT_ARGS(a,b,c,d)); C_ASSERT(5 == COUNT_ARGS(a,b,c,d,e)); C_ASSERT(6 == COUNT_ARGS(a,b,c,d,e,f)); }

 

 

转载于:https://www.cnblogs.com/shangdawei/archive/2013/06/09/3129233.html