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

Question

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?

Solution

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.

Comments

[Qix - MONICA WAS MISTREATED]

+1 Learned two things; sizeof... and constexpr. :)

[panzi]

So this sizeof... actually returns the number of arguments and not the combined storage size of all arguments (like sizeof on an array would)?

[Nawaz]

@panzi: Yes. sizeof ...(T) returns number of types packed in T. If you want to compute aggregated size of the packed types, then you've to do something like this: ideone.com/udggBk I've added this in my answer as well.

[Nawaz]

@panzi: the computation in my answer is slightly improved now.

[panzi]

@Nawaz No that is what I want. I just was suspicious because of the different behavior of sizeof. Thanks. :)

[P0W]

With C++17, to calculate size of individual arg types now we can use fold expressions return (0 + ... + sizeof(t));

[Nawaz]

@P0W: Yes. Thanks. I'll add that to my answer once C++17 is out.