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POManually releasing boost locks?
 

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    <p>For the sake of learning combinatorics of boost::thread I'm implementing a simple barrier (BR) for threads which lock a common mutex (M). However, as far as I get it when going to BR.wait() the locks on the mutex are not released, so in order for all threads to arrive at BR a manual release of the lock on M needs to be performed. So I have the following code: </p> <pre><code>boost::barrier BR(3); boost::mutex M; void THfoo(int m){ cout&lt;&lt;"TH"&lt;&lt;m&lt;&lt;" started and attempts locking M\n"; boost::lock_guard&lt;boost::mutex&gt; ownlock(M); cout&lt;&lt;"TH"&lt;&lt;m&lt;&lt;" locked mutex\n"; Wait_(15); //simple wait for few milliseconds M.unlock(); //probably bad idea //boost::lock_guard&lt;boost::mutex&gt; ~ownlock(M); // this TH needs to unlock the mutex before going to barrier BR cout&lt;&lt;"TH"&lt;&lt;m&lt;&lt;" unlocked mutex\n"; cout&lt;&lt;"TH"&lt;&lt;m&lt;&lt;" going to BR\n"; BR.wait(); cout&lt;&lt;"TH"&lt;&lt;m&lt;&lt;" let loose from BR\n"; } int main() { boost::thread TH1(THfoo,1); boost::thread TH2(THfoo,2); boost::thread TH3(THfoo,3); TH2.join(); //but TH2 might end before TH1, and so destroy BR and M cout&lt;&lt;"exiting main TH \n"; return 0; } </code></pre> <p>Whereas M.unlock() is clearly a bad solution (not using the lock); so how to (simply) release the lock? Also: how do I (properly) wait in main() for all threads to finish? (TH2.join() is bad, `cause TH2 may finish first...); </p> <p>Please do not suggest go-arounds, e.g. with conditional variables, which I can also use, but it must be possible to do it straightforwardly without them. </p>
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