Issue
I’d like to write a template function which can invoke a function with given parameters.
For instance, I can write a simple invoke function:
template<class F, class... Args>
inline auto invoke(F &&func, Args&&... args) -> decltype(auto)
{
return std::forward<F>(func)(std::forward<Args>(args)...);
}
This invoke
accepts same count of parameter which f
requires. However, I want to this template function allow additional unused parameters.
That is, I want to write some code like:
auto f = [] (auto a) {...};
invoke(f, 1, 2, 3);
Here, f
accepts only one parameter so, I wish invoke
ignore other parameters except the first one.
This can be accomplished very easily by get arity of the lambda unless the lambda is generic.
Since f
here is generic lambda, as far as I know, there’s no general way to figure out arity of f
without explicit instantiation of its template operator()<...>
.
How can I wirte my invoke
?
Solution
One possibility:
#include <utility>
#include <cstddef>
#include <tuple>
template <std::size_t... Is, typename F, typename Tuple>
auto invoke_impl(int, std::index_sequence<Is...>, F&& func, Tuple&& args)
-> decltype(std::forward<F>(func)(std::get<Is>(std::forward<Tuple>(args))...))
{
return std::forward<F>(func)(std::get<Is>(std::forward<Tuple>(args))...);
}
template <std::size_t... Is, typename F, typename Tuple>
decltype(auto) invoke_impl(char, std::index_sequence<Is...>, F&& func, Tuple&& args)
{
return invoke_impl(0
, std::index_sequence<Is..., sizeof...(Is)>{}
, std::forward<F>(func)
, std::forward<Tuple>(args));
}
template <typename F, typename... Args>
decltype(auto) invoke(F&& func, Args&&... args)
{
return invoke_impl(0
, std::index_sequence<>{}
, std::forward<F>(func)
, std::forward_as_tuple(std::forward<Args>(args)...));
}
Answered By – Piotr Skotnicki
Answer Checked By – David Goodson (BugsFixing Volunteer)