std::apply
From cppreference.com
| Defined in header <tuple>
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| template <class F, class Tuple> constexpr decltype(auto) apply(F&& f, Tuple&& t); |
(since C++17) | |
Invoke the Callable object f with a tuple of arguments.
Parameters
| f | - | Callable object to be invoked
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| t | - | tuple whose elements to be used as arguments to f
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Return value
What returned by f.
Notes
The tuple need not be std::tuple, and instead may be anything that supports std::get and std::tuple_size; in particular, std::array and std::pair may be used.
Possible implementation
namespace detail { template <class F, class Tuple, std::size_t... I> constexpr decltype(auto) apply_impl(F &&f, Tuple &&t, std::index_sequence<I...>) { return std::invoke(std::forward<F>(f), std::get<I>(std::forward<Tuple>(t))...); } } // namespace detail template <class F, class Tuple> constexpr decltype(auto) apply(F &&f, Tuple &&t) { return detail::apply_impl( std::forward<F>(f), std::forward<Tuple>(t), std::make_index_sequence<std::tuple_size_v<std::decay_t<Tuple>>>{}); } |
Example
Run this code
#include <iostream> #include <tuple> int add(int first, int second) { return first + second; } template<typename T> T add_generic(T first, T second) { return first + second; } int main() { std::cout << std::apply(add, std::make_tuple(1,2)) << '\n'; // template argument deduction/substitution fails // std::cout << std::apply(add_generic, std::make_tuple(2.0f,3.0f)) << '\n'; }
Output:
3
See also
creates a tuple object of the type defined by the argument types (function template) | |
creates a tuple of rvalue references (function template) | |
| (C++17) |
Construct an object with a tuple of arguments (function template) |
| (C++17) |
invokes any Callable object with given arguments (function template) |