std::invoke, std::invoke_r
From cppreference.com
< cpp | utility | functional
Defined in header <functional>
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(1) | ||
template< class F, class... Args > std::invoke_result_t<F, Args...> |
(since C++17) (until C++20) |
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template< class F, class... Args > constexpr std::invoke_result_t<F, Args...> |
(since C++20) | |
template< class R, class F, class... Args > constexpr R invoke_r( F&& f, Args&&... args ) noexcept(/* see below */); |
(2) | (since C++23) |
1) Invoke the Callable object
f
with the parameters args
. As by INVOKE(std::forward<F>(f), std::forward<Args>(args)...). This overload participates in overload resolution only if std::is_invocable_v<F, Args...> is true.2) Same as (1), except that the result is implicitly converted to
R
if R
is not possibly cv-qualified void, or discarded otherwise. This overload participates in overload resolution only if std::is_invocable_r_v<R, F, Args...> is true.The operation INVOKE(f, t1, t2, ..., tN) is defined as follows:
- If
f
is a pointer to member function of classT
:
- If std::is_base_of<T, std::decay_t<decltype(t1)>>::value is true, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.*f)(t2, ..., tN)
- If std::decay_t<decltype(t1)> is a specialization of std::reference_wrapper, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.get().*f)(t2, ..., tN)
- If
t1
does not satisfy the previous items, then INVOKE(f, t1, t2, ..., tN) is equivalent to ((*t1).*f)(t2, ..., tN).
- Otherwise, if N == 1 and
f
is a pointer to data member of classT
:
- If std::is_base_of<T, std::decay_t<decltype(t1)>>::value is true, then INVOKE(f, t1) is equivalent to t1.*f
- If std::decay_t<decltype(t1)> is a specialization of std::reference_wrapper, then INVOKE(f, t1) is equivalent to t1.get().*f
- If
t1
does not satisfy the previous items, then INVOKE(f, t1) is equivalent to (*t1).*f
- Otherwise, INVOKE(f, t1, t2, ..., tN) is equivalent to f(t1, t2, ..., tN) (that is,
f
is a FunctionObject)
Parameters
f | - | Callable object to be invoked |
args | - | arguments to pass to f
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Return value
1) The value returned by
f
.2) The value returned by
f
, implicitly converted to R
, if R
is not void. None otherwise.Exceptions
1)
noexcept specification:
noexcept(std::is_nothrow_invocable_v<F, Args...>)
2)
noexcept specification:
noexcept(std::is_nothrow_invocable_r_v<R, F, Args...>)
Possible implementation
First version |
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namespace detail { template<class> constexpr bool is_reference_wrapper_v = false; template<class U> constexpr bool is_reference_wrapper_v<std::reference_wrapper<U>> = true; template<class C, class Pointed, class T1, class... Args> constexpr decltype(auto) invoke_memptr(Pointed C::* f, T1&& t1, Args&&... args) { if constexpr (std::is_function_v<Pointed>) { if constexpr (std::is_base_of_v<C, std::decay_t<T1>>) return (std::forward<T1>(t1).*f)(std::forward<Args>(args)...); else if constexpr (is_reference_wrapper_v<std::decay_t<T1>>) return (t1.get().*f)(std::forward<Args>(args)...); else return ((*std::forward<T1>(t1)).*f)(std::forward<Args>(args)...); } else { static_assert(std::is_object_v<Pointed> && sizeof...(args) == 0); if constexpr (std::is_base_of_v<C, std::decay_t<T1>>) return std::forward<T1>(t1).*f; else if constexpr (is_reference_wrapper_v<std::decay_t<T1>>) return t1.get().*f; else return (*std::forward<T1>(t1)).*f; } } } // namespace detail template<class F, class... Args> constexpr std::invoke_result_t<F, Args...> invoke(F&& f, Args&&... args) noexcept(std::is_nothrow_invocable_v<F, Args...>) { if constexpr (std::is_member_pointer_v<std::decay_t<F>>) return detail::invoke_memptr(f, std::forward<Args>(args)...); else return std::forward<F>(f)(std::forward<Args>(args)...); } |
Second version |
template<class R, class F, class... Args> requires std::is_invocable_r_v<R, F, Args...> constexpr R invoke_r(F&& f, Args&&... args) noexcept(std::is_nothrow_invocable_r_v<R, F, Args...>) { if constexpr (std::is_void_v<R>) std::invoke(std::forward<F>(f), std::forward<Args>(args)...); else return std::invoke(std::forward<F>(f), std::forward<Args>(args)...); } |
Notes
Feature-test macro: | __cpp_lib_invoke (for std::invoke )
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Feature-test macro: | __cpp_lib_invoke_r (for std::invoke_r )
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Example
Run this code
#include <functional> #include <iostream> #include <type_traits> struct Foo { Foo(int num) : num_(num) {} void print_add(int i) const { std::cout << num_+i << '\n'; } int num_; }; void print_num(int i) { std::cout << i << '\n'; } struct PrintNum { void operator()(int i) const { std::cout << i << '\n'; } }; int main() { // invoke a free function std::invoke(print_num, -9); // invoke a lambda std::invoke([]() { print_num(42); }); // invoke a member function const Foo foo(314159); std::invoke(&Foo::print_add, foo, 1); // invoke (access) a data member std::cout << "num_: " << std::invoke(&Foo::num_, foo) << '\n'; // invoke a function object std::invoke(PrintNum(), 18); # if defined(__cpp_lib_invoke_r) auto add = [](int x, int y) { return x + y; }; auto ret = std::invoke_r<float>(add, 11, 22); static_assert(std::is_same<decltype(ret), float>()); std::cout << ret << '\n'; std::invoke_r<void>(print_num, 44); # endif }
Possible output:
-9 42 314160 num_: 314159 18 33 44
See also
(C++11) |
creates a function object out of a pointer to a member (function template) |
(C++11)(removed in C++20)(C++17) |
deduces the result type of invoking a callable object with a set of arguments (class template) |
checks if a type can be invoked (as if by std::invoke) with the given argument types (class template) | |
(C++17) |
calls a function with a tuple of arguments (function template) |