c++ - How to extract __VA_ARGS__? -

i hava macro call static function each args.

for example:

#define foo(x) x::do(); #define foo_1(x,y) x::do(); y::do(); 

my question need use foo variable number of arguments, possible use __va_args__ ?

like line below:

#define foo(...) __va_args__::do() ?  

thanks

macro expansion not work argument pack expansion variadic templates. have expand to:

x,y::do(); 

and not to

x::do(); y::do(); 

as hoped. in c++11 use variadic templates. instance, want way:

#include <iostream>  struct x { static void foo() { std::cout << "x::foo()" << std::endl; }; }; struct y { static void foo() { std::cout << "y::foo()" << std::endl; }; }; struct z { static void foo() { std::cout << "z::foo()" << std::endl; }; };  int main() {     do_foo<x, y, z>(); } 

all need (relatively simple) machinery:

namespace detail {     template<typename... ts>     struct do_foo;      template<typename t, typename... ts>     struct do_foo<t, ts...>     {         static void call()         {             t::foo();             do_foo<ts...>::call();         }     };      template<typename t>     struct do_foo<t>     {         static void call()         {             t::foo();         }     }; }  template<typename... ts> void do_foo() {     detail::do_foo<ts...>::call(); } 

here live example.