Jacques Kerner said:
> Hi,
>
> I get the following error :
>
> error C2664: 'void boost::function0<R,Allocator>::assign_to(Functor)' :
> unable to convert parameter 1 from 'const CTaskManager' to
> 'CTaskManager'
>
> when doing this :
>
> class CTaskManager
> {
> public:
> CTaskManager();
> ~CTaskManager();
> void operator()() {}
>
> private:
> boost::mutex m_mutex;
> }
>
> and
>
> CTaskManager taskManager;
> boost::thread_group mainThreadGroup;
> mainThreadGroup.create_thread(taskManager);
> mainThreadGroup.join_all();
>
> The error dissapears when I remove the mutex from the definition of
> CTaskManager ... (?!!)
Correct. Functors are passed by value (i.e. they must be Copyable), and
Mutexes are Noncopyable.
> So what is the right way to use mutex and threads together? Do I have to
> declare the mutex outside of the
> functor? Why?
No, you just have to enable the functor to be copyable, as per the FAQ at
http://www.boost.org/libs/thread/doc/faq.html#question5.
However, I'm going to guess from the code snippet that you really don't
want this functor to be copied? If that's the case, you may want to make
use of boost::ref.
CTaskManager taskManager;
boost::thread_group mainThreadGroup;
mainThreadGroup.create_thread(boost::ref(taskManager));
mainThreadGroup.join_all();
--
William E. Kempf
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