C + + 11: 可变参数模板函数参数个数? ？

如何计算可变参数模板函数的参数个数？

即:

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


...
}

实现上述 number_of_args的最佳方法是什么？

44577 次浏览
小开
最佳答案

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.

小开 
#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;
}

提交答案