std::invoke, std::invoke_r
From cppreference.com
< cpp | utility | functional
| Defined in header <functional>
|
||
template< class F, class... Args > std::invoke_result_t<F, Args...> |
(1) | (since C++17) (constexpr since C++20) |
template< class R, class F, class... Args > constexpr R |
(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) Invoke the Callable object f with the parameters args as by
INVOKE<R>(std::forward<F>(f), std::forward<Args>(args)...). This overload participates in overload resolution only if std::is_invocable_r_v<R, F, Args...> is true.Contents |
[edit] Parameters
| f | - | Callable object to be invoked |
| args | - | arguments to pass to f |
[edit] Return value
1) The value returned by f.
2) The value returned by f, implicitly converted to
R, if R is not (possibly cv-qualified) void. None otherwise.[edit] 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...>)
[edit] Possible implementation
| invoke (1) |
|---|
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 T> using remove_cvref_t = std::remove_cv_t<std::remove_reference_t<T>>; template<class C, class Pointed, class Object, class... Args> constexpr decltype(auto) invoke_memptr(Pointed C::* member, Object&& object, Args&&... args) { using object_t = remove_cvref_t<Object>; constexpr bool is_member_function = std::is_function_v<Pointed>; constexpr bool is_wrapped = is_reference_wrapper_v<object_t>; constexpr bool is_derived_object = std::is_same_v<C, object_t> || std::is_base_of_v<C, object_t>; if constexpr (is_member_function) { if constexpr (is_derived_object) return (std::forward<Object>(object) .* member) (std::forward<Args>(args)...); else if constexpr (is_wrapped) return (object.get() .* member)(std::forward<Args>(args)...); else return ((*std::forward<Object>(object)) .* member) (std::forward<Args>(args)...); } else { static_assert(std::is_object_v<Pointed> && sizeof...(args) == 0); if constexpr (is_derived_object) return std::forward<Object>(object) .* member; else if constexpr (is_wrapped) return object.get() .* member; else return (*std::forward<Object>(object)) .* member; } } } // 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<detail::remove_cvref_t<F>>) return detail::invoke_memptr(f, std::forward<Args>(args)...); else return std::forward<F>(f)(std::forward<Args>(args)...); } |
| invoke_r (2) |
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)...); } |
[edit] Notes
| Feature-test macro | Value | Std | Feature |
|---|---|---|---|
__cpp_lib_invoke |
201411L | (C++17) | std::invoke, (1)
|
__cpp_lib_invoke_r |
202106L | (C++23) | std::invoke_r, (2)
|
[edit] 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() { std::cout << "invoke a free function: "; std::invoke(print_num, -9); std::cout << "invoke a lambda: "; std::invoke([](){ print_num(42); }); std::cout << "invoke a member function: "; const Foo foo(314159); std::invoke(&Foo::print_add, foo, 1); std::cout << "invoke (i.e., access) a data member num_: " << std::invoke(&Foo::num_, foo) << '\n'; std::cout << "invoke a function object: "; std::invoke(PrintNum(), 18); #if defined(__cpp_lib_invoke_r) auto add = [](int x, int y){ return x + y; }; std::cout << "invoke a lambda converting result to float: "; auto ret = std::invoke_r<float>(add, 11, 22); static_assert(std::is_same<decltype(ret), float>()); std::cout << std::fixed << ret << "\ninvoke print_num: "; std::invoke_r<void>(print_num, 44); #endif }
Possible output:
invoke a free function: -9 invoke a lambda: 42 invoke a member function: 314160 invoke (i.e., access) a data member num_: 314159 invoke a function object: 18 invoke a lambda converting result to float: 33.000000 invoke print_num: 44
[edit] 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) |
