Re: [boost] boost::signal thread safety (implied question: is boost::function thread safe ?)

From: Tanguy Fautre (tanguy.fautre_at_[hidden])
Date: 2007-03-02 05:47:51

Stefan Seefeld wrote:
> Philippe Vaucher wrote:
>> Hello,
>>
>> Well, the subject says it all ;)
>> This question comes because boost::signal's documentation says it's not
>> thread safe, but I have trouble finding exactly why it isn't.
>>
>> Let's look the following example:
>>
>> Each thread is connected to the same signal, then the signal is emmited, and
>> each threads receives the message.... Why would this not be thread safe ?
> 
> You use 'thread' in a very lose way here. There is a 'thread' as seen by the
> OS, i.e. an execution entity processing some code, and there is are two
> 'thread' types ('thread1' and 'thread2') you use.
> 
> In your description above you seem to refer to those types as 'threads',
> not to the execution entities.
> 
> The point is, the 'threads of execution' don't 'receive the message', as
> the 'slot()' methods are executed in the main thread. However, as you check
> the value of 'finished' in other 'threads of execution', you have to protect
> access to it by a mutex. For example:
> 
> struct thread1
> {
>   thread1() : finished_(false) {}
>   void operator()()
>   {
>     while (true)
>     {
>       scoped_lock lock(mutex_);
>       if (finished_) return;
>     }
>   }
>   void slot()
>   {
>     scoped_lock lock(mutex_);
>     finished_ = true;
>   }
> 
>   bool fininshed_;
>   mutex mutex_;
> };
> 
> 
> HTH,
> 		Stefan
> 

I'm not sure whether a mutex is really needed here. Because a boolean is 
a native type, I don't see any particular problem if one thread reads it 
while it's being written by another.
So I think the mutex can be safely removed.

What I would however do is to use "volatile bool" instead of just "bool"

Tanguy