Reputation: 68618
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?
Upvotes: 108
Views: 57556
Reputation: 361322
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) + ...);
Upvotes: 131
Reputation: 41
#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;
}
Upvotes: 3