如何在boost :: unique_lock上尝试try_lock <boost::mutex>

Question

As per title, how to try_lock on a boost::unique_lock ?

I've this code:

void mySafeFunct()
{
    if(myMutex.try_lock() == false)
    {
        return -1;
    }

    // mutex ownership is automatically acquired

    // do stuff safely

    myMutex.unlock();
}

Now I'd like to use a unique_lock (which is also a scoped mutex) instead of the plain boost::mutex. I want this to avoid all the unlock() calls from the function body.

Answer 1

You can either defer the locking with the Defer constructor , or use the the Try constructor when creating your unique_lock :

boost::mutex myMutex;
boost::unique_lock<boost::mutex> lock(myMutex, boost::try_lock);

if (!lock.owns_lock())
    return -1;

...

Answer 2

boost::mutex myMutex;
boost::unique_lock<boost::mutex> lock(myMutex, boost::defer_lock);
lock.try_lock()

Answer 3

Previous answers may be outdated. I'm using boost 1.53 and this seems to work:

boost::unique_lock<boost::mutex> lk(myMutex, boost::try_to_lock);
if (lk)
  doTheJob();