C++11：可变模板函数参数的数量？
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C++11: Number of Variadic Template Function Parameters?
提问by Andrew Tomazos
How can I get a count of the number of arguments to a variadic template function?
如何获取可变参数模板函数的参数数量？
ie:
IE：
template<typename... T>
void f(const T&... t)
{
int n = number_of_args(t);
...
}
What is the best way to implement number_of_argsin the above?
number_of_args在上面实施的最佳方法是什么？
回答by Nawaz
Just write this:
只写这个：
const std::size_t n = sizeof...(T); //you may use `constexpr` instead of `const`
Note that nis a constant expression (i.e known at compile-time), which means you may use it where constant expression is needed, such as:
请注意，这n是一个常量表达式（即在编译时已知），这意味着您可以在需要常量表达式的地方使用它，例如：
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 numberof types in the pack), then you've to do something like this:
请注意，如果您想计算打包类型的聚合大小（而不是包中的类型数量），那么您必须执行以下操作：
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:
然后这样做：
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 foldexpression:
在 C++17（及更高版本）中，使用fold表达式计算类型的大小总和要简单得多：
constexpr auto size = (sizeof(T) + ...);
Hope that helps.
希望有帮助。
