Why is std::packaged_task not valid?

Question

Using MSVC2012,

The following code will compile and run as expected

std::packaged_task< int() > task( []()->int{ std::cout << \"hello          


        

Answer 1

The problem is still there in MSVS 2013RC, but I made this temporary patch while MS corrects it. It is a specialization of packaged_task for void(...), so I suggest putting this in a header file and including it after the standard headers.Note that make_ready_at_thread_exit() is not implemented and some functions have not been fully tested, use at your own risk.

namespace std {

template
class packaged_task
{
    promise _my_promise;
    function _my_func;

public:
    packaged_task() {
    }

    template
    explicit packaged_task(_Fty2&& _Fnarg)
      : _my_func(_Fnarg) {
    }

    packaged_task(packaged_task&& _Other)
      : _my_promise(move(_Other._my_promise)),
        _my_func(move(_Other._my_func)) {
    }

    packaged_task& operator=(packaged_task&& _Other) {
        _my_promise = move(_Other._my_promise);
        _my_func = move(_Other._my_func);
        return (*this);
    }

    packaged_task(const packaged_task&) = delete;
    packaged_task& operator=(const packaged_task&) = delete;

    ~packaged_task() {
    }

    void swap(packaged_task& _Other) {
        _my_promise.swap(_Other._my_promise);
        _my_func.swap(_Other._my_func);
    }

    explicit operator bool() const {
        return _my_func != false;
    }

    bool valid() const {
        return _my_func != false;
    }

    future get_future() {
        return _my_promise.get_future();
    }

    void operator()(_ArgTypes... _Args) {
        _my_func(forward<_ArgTypes>(_Args)...);
        _my_promise.set_value();
    }

    void reset() {
        swap(packaged_task());
    }
};

}; // namespace std