C++ preprocessor __VA_ARGS__ number of arguments

Question

Simple question for which I could not find answer on the net. In variadic argument macros, how to find the number of arguments? I am okay with boost preprocessor, if it has

Answer 1

this works with 0 arguments with gcc/llvm. [links are dumb]

/*
 * we need a comma at the start for ##_VA_ARGS__ 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_(,##__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));
   return 0;
}

Visual Studio seems to be ignoring the ## operator used to consume the empty argument. You can probably get around that with something like

#define CNT_ COUNT_ARGS
#define PASTE(x,y) PASTE_(x,y)
#define PASTE_(x,y) x ## y
#define CNT(...) PASTE(ARGVS,PASTE(CNT_(__VA_ARGS__),CNT_(1,##__VA_ARGS__)))
//you know its 0 if its 11 or 01
#define ARGVS11 0
#define ARGVS01 0
#define ARGVS12 1
#define ARGVS23 2
#define ARGVS34 3