c++11模板类返回类型

Question

一个简单的类

   struct Test{
        void display(int x)
        {
           printf("%d\n", x);
        }
    };

c++11

template<class F, class... Args>
auto enqueue(F&& f, Args&&... args) -> std::future<decltype(f(args...))>{

    auto task = std::make_shared<std::packaged_task<decltype(f(args...)())>>(
        std::bind(std::forward<F>(f), std::forward<Args>(args)...)
    );
    return task->get_future();
}

称之为just use (cls是上面模板函数类的对象)

Test test;
cls->enqueue(&Test::display, &test, 0);

计算器错误

candidate template ignored: substitution failure 
[with F = void (TestMem::*)(int), Rest = <TestMem *, int>]: called object type 'void (TestMem::*)(int)' is not a function or function pointer
        auto enqueue(F && f, Rest&&... rest) ->std::future<decltype(f(rest...))> {

上面在非类成员函数上工作得很好，有人能给c++11一个修复吗？

c++14运行良好，但我的项目需要c++11

template<class F, class... Args>
auto enqueue(F&& f, Args&&... args) {

    using return_type = typename std::result_of<F(Args...)>::type;

    auto task = std::make_shared<std::packaged_task<return_type()>>(
        std::bind(std::forward<F>(f), std::forward<Args>(args)...)
    );
    return task->get_future();
}

Answer 1

将返回类型定义为std::future::type>解决此问题