----- Original Message ----- From: "Anthony Williams" <anthony.ajw@gmail.com> To: <boost@lists.boost.org> Sent: Tuesday, February 10, 2009 6:28 PM Subject: Re: [boost] Futures Review - Should packaged_taskbeCopyConstructible?
"vicente.botet" <vicente.botet@wanadoo.fr> writes:
So I've
tried to move the packaged_task which don't work either because the promise is broken
template<typename F> std::packaged_task<typename std::result_of<F()>::type> lazy_call(F f) { typedef typename std::result_of<F()>::type result_type; std::packaged_task<result_type> task(std::move(f)); task.set_wait_callback(invoke_lazy_task<result_type>()) return task; }
Why does this not work? Is it that the callback is broken (it gets passed the wrong task?) If so, that's the part that needs fixing, rather than the copyability of packaged_task.
Sorry i was not clear. The problem is that the reference to the task std::packaged_task<result_type> task(std::move(f)); has been stored in the set_wait_callback. But this address is no more valable when the function returns, so when the user gets the future and call to get, the callback uses a bad address. Hoping it is clear now.
What is the rationale for making packaged_task movable only? Why should not be CopyConstructible?
If packaged_task was copyable, people would copy it and expect the copy to be associated with the futures they obtained from the original. Not necessarily deliberately, but by mistake (e.g. copying into a vector rather than moving). This is not the case.
Ok, I see. Anyway, I think that examples like template<typename F> std::packaged_task<typename std::result_of<F()>::type> lazy_call(F f); should be included in the library, at least as an example. Best, Vicente