c++ - Variadic macro argument count not working as expected -

so, i'm trying implement macro count number of arguments in va_args.

for sake of simplicity works 3 parameters. problem when macro used less 3 parameters, doesn't work, , triggers "expected expression" error.

#define expand( x ) x #define pp_narg(...) expand(pp_arg_n(__va_args__, pp_rseq_n())) #define pp_arg_n(_1, _2, _3, n,...) n #define pp_rseq_n() 3,2,1,0  void main() {     printf("\ntest pp_narg: %i", pp_narg());        //doesn't work (in case shouldn't work, it's correct)     printf("\ntest pp_narg: %i", pp_narg(0));       //doesn't work     printf("\ntest pp_narg: %i", pp_narg(0,0));     //doesn't work     printf("\ntest pp_narg: %i", pp_narg(0,0,0));   //works } 

keeping line works compiles correctly , prints "test pp_narg: 3".

i believe problem might pp_rseq_n() expanding "3", instead of "3,2,1,0" reason, since if pp_rseq_n() defined this

#define pp_rseq_n() 10,9,8,7,6,5,4,3,2,1,0 

it still doesn't work less 3 parameters.

im using msvc compiler , may cause of problem, since doesn't behave macros, seen here: msvc doesn't expand __va_args__ correctly

in implementation pp_rseq_n() argument pp_arg_n. argument, expanded in argument substitution phase of preprocessing, happens prior replacing argument in replacement list (so long as, in replacement list, it's not being stringified , not participating in paste).

since pp_arg_n has fourth argument n in replacement list, pp_rseq_n() expand if happen pass 3 arguments in. (there second scan during rescan , replacement phase, applies after argument substitution... has no effect here pp_rseq_n() mentioned in call).

get rid of macro , put in pp_narg this:

#define expand( x ) x #define pp_narg(...) expand(pp_arg_n(__va_args__, 3,2,1,0)) #define pp_arg_n(_1, _2, _3, n,...) n 

...and things "work" fine:

pp_narg() expands 1
pp_narg(x) expands 1
pp_narg(x,y) expands 2

note, however, pp_narg() doesn't give 0. arguably that's correct; preprocessor, not passing 0 arguments. it's passing 1 argument empty. it's same thing #define x(a) open close/x() yielding open close. if reason want expand 0, there may finagling happen, answer i'm focusing on getting on 1 hump.