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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