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std:: invoke, std:: invoke_r

From cppreference.net
C++
Utilities library
Function objects
Function invocation
invoke invoke_r
(C++17) (C++23)
Identity function object
(C++20)
Old binders and adaptors
( until C++17* )
( until C++17* )
( until C++17* )
( until C++17* )
( until C++17* )
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( until C++17* ) ( until C++17* ) ( until C++17* ) ( until C++17* )
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( until C++17* )
( until C++17* ) ( until C++17* ) ( until C++17* ) ( until C++17* )
( until C++20* )
( until C++20* )
定义于头文件 <functional>
template < class F, class ... Args >

std:: invoke_result_t < F, Args... >

invoke ( F && f, Args && ... args ) noexcept ( /* see below */ ) ;
(1) (C++17 起)
(C++20 起为 constexpr)
template < class R, class F, class ... Args >

constexpr R

invoke_r ( F && f, Args && ... args ) noexcept ( /* see below */ ) ;
(2) (C++23 起)
1) 通过 INVOKE ( std:: forward < F > ( f ) , std:: forward < Args > ( args ) ... ) 的方式调用 Callable 对象 f 并传递参数 args 。此重载仅当 std:: is_invocable_v < F, Args... > 为 true 时参与重载决议。
2) 通过 INVOKE<R> ( std:: forward < F > ( f ) , std:: forward < Args > ( args ) ... ) 的方式调用 Callable 对象 f 并传递参数 args 。此重载仅当 std:: is_invocable_r_v < R, F, Args... > 为 true 时参与重载决议。

目录

参数

f - 待调用的 Callable 对象
args - 传递给 f 的参数

返回值

1) 由 f 返回的值。
2) 当 R 不是（可能被 cv限定 的） void 时，返回由 f 返回并隐式转换为 R 的值。否则不返回任何值。

异常

1)
noexcept 规范：
noexcept ( std:: is_nothrow_invocable_v < F, Args... > )
2)
noexcept 规范：
noexcept ( std:: is_nothrow_invocable_r_v < R, F, Args... > )

可能的实现

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<span class="br

注释

功能测试 宏 值 标准 功能
__cpp_lib_invoke 201411L (C++17) std::invoke , ( 1 )
__cpp_lib_invoke_r 202106L (C++23) std::invoke_r , ( 2 )

示例

运行此代码 
#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
}

可能的输出： 

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

参见

(C++11)
从成员指针创建函数对象
(函数模板)
(C++11) (removed in C++20) (C++17)
推导调用可调用对象时带一组实参的结果类型
(类模板)
(C++17)
检查类型是否能以给定实参类型调用（如同以 std::invoke ）
(类模板)
(C++17)
以实参的元组调用函数
(函数模板)