I am trying to learn about variadic templates in C++11. I have a class which is basically a wrapper around a std::array
. I want to be able to pass function objects (ideally lambdas) to a member function and then have the elements of the std::array
passed on as parameters of the function object.
I have used a static_assert
to check that the number of parameters matches the length of the array but I cannot think of a way to pass the elements as arguments.
Here is the code
#include <iostream>
#include <array>
#include <memory>
#include <initializer_list>
using namespace std;
template<int N, typename T>
struct Container {
template<typename... Ts>
Container(Ts&&... vs) : data{{std::forward<Ts>(vs)...}} {
static_assert(sizeof...(Ts)==N,"Not enough args supplied!");
}
template< typename... Ts>
void doOperation( std::function<void(Ts...)>&& func )
{
static_assert(sizeof...(Ts)==N,"Size of variadic template args does not match array length");
// how can one call func with the entries
// of data as the parameters (in a way generic with N)
}
std::array<T,N> data;
};
int main(void)
{
Container<3,int> cont(1,2,3);
double sum = 0.0;
auto func = [&sum](int x, int y, int z)->void{
sum += x;
sum += y;
sum += z;
};
cont.doOperation(std::function<void(int,int,int)>(func));
cout << sum << endl;
return 0;
}
So my question (as indicated in the code) is how can one pass the entries of data
onto the function func
in a way which is generic with N
?
Bonus Question: Is it possible to do away with the unsightly conversion to std::function
in main and pass in a lambda directly?
Given the well-known indices infrastructure:
namespace detail
{
template<int... Is>
struct seq { };
template<int N, int... Is>
struct gen_seq : gen_seq<N - 1, N - 1, Is...> { };
template<int... Is>
struct gen_seq<0, Is...> : seq<Is...> { };
}
You could redefine your class template this way:
template<int N, typename T>
struct Container {
template<typename... Ts>
Container(Ts&&... vs) : data{{std::forward<Ts>(vs)...}} {
static_assert(sizeof...(Ts)==N,"Not enough args supplied!");
}
template<typename F>
void doOperation(F&& func)
{
doOperation(std::forward<F>(func), detail::gen_seq<N>());
}
template<typename F, int... Is>
void doOperation(F&& func, detail::seq<Is...>)
{
(std::forward<F>(func))(data[Is]...);
}
std::array<T,N> data;
};
Here is a live example .
Notice, that you do not need to construct an std::function
object in main()
: the std::function
can be implicitly constructed from the lambda. However, you do not even need to use std::function
at all here, possibly incurring in an unnecessary run-time overhead.
In the solution above, I just let the type of the callable object to be a template parameter that can be deduced by the compiler.
You can use this utility template to create a sequence of indices at compile time:
template< std::size_t... Ns >
struct indices {
typedef indices< Ns..., sizeof...( Ns ) > next;
};
template< std::size_t N >
struct make_indices {
typedef typename make_indices< N - 1 >::type::next type;
};
template<>
struct make_indices< 0 > {
typedef indices<> type;
};
Then make a caller function that takes indices
as a parameter so you got a way of deducing the indices sequence:
template<typename... Ts, size_t...Is>
void call(std::function<void(Ts...)>&& func, indices<Is...>)
{
func( data[Is]... );
}
And then you can call it like this:
template< typename... Ts>
void doOperation( std::function<void(Ts...)>&& func )
{
static_assert(sizeof...(Ts)==N,"Size of variadic template args does not match array length");
call( std::forward<std::function<void(Ts...)>>(func), typename make_indices<N>::type() );
}
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