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std::apply

From cppreference.com
< cpp‎ | utility
Defined in header <tuple>
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.

Contents

[edit] Parameters

f - Callable object to be invoked
t - tuple whose elements to be used as arguments to f

[edit] Return value

What returned by f.

[edit] 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.

[edit] 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>>>{});
}

[edit] Example

#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

[edit] See also

creates a tuple object of the type defined by the argument types
(function template) [edit]
creates a tuple of rvalue references
(function template) [edit]
Construct an object with a tuple of arguments
(function template) [edit]
(C++17)
invokes any Callable object with given arguments
(function template) [edit]