C ++ 17结合了std :: async和std :: invoke

Question

I am trying to call std::invoke from inside a std::async call but for some reason the compiler don't like it.

Note : I know I could use a lambda, but I would like to make it work without it

Here is my example.

#include <iostream>
#include <future>
#include <string>
#include <functional>
#include <type_traits>
#include <unistd.h>

template <class Fn, class... Args>
inline std::result_of_t<Fn&&(Args&&...)> runTerminateOnException(Fn&& fn, Args&&... args) {
    try {
        return std::invoke(std::forward<Fn>(fn), std::forward<Args>(args)...);
    } catch (...) {
        std::terminate();
    }
}

struct A {
        static void g(double x, std::string *s) {
            std::cout << "g() : x = " << x << ", *s = " << *s << std::endl;
            usleep(100);
        }

        static void f(double x, std::string *s) {
            std::invoke(g, x, s); // Working
            auto future1 = std::async(std::launch::async, g, x, s); // Working
            auto future2 = std::async(std::launch::async, std::invoke, g, x, s); // Not working
            auto future3 = std::async(std::launch::async, runTerminateOnException, g, x, s); // Not working
        }
};

int main() {
    std::string s = "Hello";
    A::f(10., &s);
    return 0;
}

Thank you for your help.

http://coliru.stacked-crooked.com/a/51f8c80b0a05fa0e

Answer 1

std::invoke is a template function. Therefore simply naming the template name is ambiguous - which of the infinite set of std::invoke<F, Args...> do you mean?

You'll need to provide an 'invoker' concrete object.

eg:

#include <iostream>
#include <future>
#include <string>
#include <functional>
#include <type_traits>
#include <unistd.h>

template <class Fn, class... Args>
inline std::result_of_t<Fn&&(Args&&...)> runTerminateOnException(Fn&& fn, Args&&... args) {
    try {
        return std::invoke(std::forward<Fn>(fn), std::forward<Args>(args)...);
    } catch (...) {
        std::terminate();
    }
}

struct invoker
{
    template<class F, class...Args>
    decltype(auto) operator()(F&& f, Args&&...args) const {
        return std::invoke(f, std::forward<Args>(args)...);
    }
};

struct A {
        static void g(double x, std::string *s) {
            std::cout << "g() : x = " << x << ", *s = " << *s << std::endl;
            usleep(100);
        }

        static void f(double x, std::string *s) {
            std::invoke(g, x, s); // Working
            auto future1 = std::async(std::launch::async, g, x, s); // Working
            auto future2 = std::async(std::launch::async, invoker(), g, x, s); // Working now
//          auto future3 = std::async(std::launch::async, runTerminateOnException, g, x, s); // Not working
        }
};

int main() {
    std::string s = "Hello";
    A::f(10., &s);
    return 0;
}

Same with the template function runTerminateOnException.