In C ++ 11, ... is it considered an operator?

Question

I was wondering if ... was considered as an operator in C ++ 11. And if so, what is its priority?

For example, consider this rather bad example and assume that ... is an operator.

template<typename T, typename...Args>
void foo(T _elm, Args... _args)
{
   bar(something,_args...);
}

How do I know whether to barits first argument, which is something, and args...expanded, or if it will be launched on the result operator,(something, _args...)? (bonus question: can operators be overloaded with variation patterns?)

Answer 1

I was wondering if ... the operator in C ++ 11 was considered

, ... ++ 11. ,

catch(...)

, ... , .

?

, . , :

int operator + (int param1, my_obj param2);

, .