C++11: Number of Variadic Template Function Parameters?

How can I get a count of the number of arguments to a variadic template function?

ie:

template<typename... T>
void f(const T&... t)
{
    int n = number_of_args(t);

    ...
}

What is the best way to implement number_of_args in the above?

Just write this:

const std::size_t n = sizeof...(T); //you may use `constexpr` instead of `const`

Note that n is a constant expression (i.e known at compile-time), which means you may use it where constant expression is needed, such as:

std::array<int,   n>  a; //array of  n elements
std::array<int, 2*n>  b; //array of (2*n) elements

auto middle = std::get<n/2>(tupleInstance);

---

Note that if you want to compute aggregated size of the packed types (as opposed to number of types in the pack), then you've to do something like this:

template<std::size_t ...>
struct add_all : std::integral_constant< std::size_t,0 > {};
 
template<std::size_t X, std::size_t ... Xs>
struct add_all<X,Xs...> : 
  std::integral_constant< std::size_t, X + add_all<Xs...>::value > {};

then do this:

constexpr auto size = add_all< sizeof(T)... >::value;

In C++17 (and later), computing the sum of size of the types is much simpler using fold expression:

constexpr auto size = (sizeof(T) + ...);

Hope that helps.

#include <iostream>

template<typename ...Args>
struct SomeStruct
{
    static const int size = sizeof...(Args);
};

template<typename... T>
void f(const T&... t)
{
    // this is first way to get the number of arguments
    constexpr auto size = sizeof...(T);
    std::cout<<size <<std::endl;
}

int main ()
{
    f("Raje", 2, 4, "ASH");
    // this is 2nd way to get the number of arguments
    std::cout<<SomeStruct<int, std::string>::size<<std::endl;
    return 0;
}