[英]Manually releasing boost locks?
为了学习boost :: thread的组合,我正在为线程实现一个简单的屏障(BR)来锁定一个普通的互斥锁(M)。 但是,就我转到BR.wait()而言,互斥锁上的锁定没有释放,因此为了让所有线程都到达BR,需要手动释放M上的锁定。 所以我有以下代码:
boost::barrier BR(3);
boost::mutex M;
void THfoo(int m){
cout<<"TH"<<m<<" started and attempts locking M\n";
boost::lock_guard<boost::mutex> ownlock(M);
cout<<"TH"<<m<<" locked mutex\n";
Wait_(15); //simple wait for few milliseconds
M.unlock(); //probably bad idea
//boost::lock_guard<boost::mutex> ~ownlock(M);
// this TH needs to unlock the mutex before going to barrier BR
cout<<"TH"<<m<<" unlocked mutex\n";
cout<<"TH"<<m<<" going to BR\n";
BR.wait();
cout<<"TH"<<m<<" 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<<"exiting main TH \n";
return 0;
}
而M.unlock()显然是一个糟糕的解决方案(不使用锁); 那么如何(简单地)释放锁? 另外:我如何(正确)在main()中等待所有线程完成? (TH2.join()很糟糕,因为TH2可能先完成......);
请不要建议复飞,例如使用条件变量,我也可以使用它,但必须可以在没有条件变量的情况下直接进行。
除了在块中确定boost::lock_guard
范围之外,您还可以使用boost::unique_lock
,它可以显式unlock()
:
boost::unique_lock<boost::mutex> ownlock(M);
cout<<"TH"<<m<<" locked mutex\n";
Wait_(15); //simple wait for few milliseconds
ownlock.unlock();
如果您需要在以后重新获取互斥锁之前释放互斥锁,这将非常有用。
至于连接,只需依次调用所有线程句柄上的join()
。
就像是:
void THfoo(int m){
// use a scope here, this means that the lock_guard will be destroyed (and therefore mutex unlocked on exiting this scope
{
cout<<"TH"<<m<<" started and attempts locking M\n";
boost::lock_guard<boost::mutex> ownlock(M);
cout<<"TH"<<m<<" locked mutex\n";
Wait_(15); //simple wait for few milliseconds
}
// This is outside of the lock
cout<<"TH"<<m<<" unlocked mutex\n";
cout<<"TH"<<m<<" going to BR\n";
BR.wait();
cout<<"TH"<<m<<" let loose from BR\n";
}
至于等待,只需在所有线程句柄上调用join(如果它们已经完成,函数将立即返回)
TH1.join();
TH2.join();
TH3.join();
让它超出范围:
void THfoo(int m){
cout<<"TH"<<m<<" started and attempts locking M\n";
{
boost::lock_guard<boost::mutex> ownlock(M);
cout<<"TH"<<m<<" locked mutex\n";
Wait_(15); //simple wait for few milliseconds
}
// this TH needs to unlock the mutex before going to barrier BR
cout<<"TH"<<m<<" unlocked mutex\n";
cout<<"TH"<<m<<" going to BR\n";
BR.wait();
cout<<"TH"<<m<<" let loose from BR\n";
}
cout<<"TH"<<m<<" started and attempts locking M\n";
{
boost::lock_guard<boost::mutex> ownlock(M);
cout<<"TH"<<m<<" locked mutex\n";
Wait_(15); //simple wait for few milliseconds
} //boost::lock_guard<boost::mutex> ~ownlock(M);
// this TH needs to unlock the mutex before going to barrier BR
cout<<"TH"<<m<<" unlocked mutex\n";
只要您join
所有线程,TH2首先完成时唯一的问题是TH1必须在TH2可以通过join
“收获”之前完成运行,并且释放任何剩余资源(如返回值)。 3线程并不值得担心。 如果该内存使用是一个问题,那么你可以使用timed_join
反复尝试所有线程。
您也可以执行您不想要的操作 - 让主线程在条件变量上等待,并且每个线程在完成时在某处存储一个值以表示它已完成,并发出条件变量的信号,以便主线程可以join
它。 你必须绝对确定线程会发出信号,否则你可能会永远等待它。 如果你取消线程,要小心。
如果你使用boost :: mutex :: scoped_lock而不是boost :: lock_guard,它有一个unlock()方法。 如果你调用它,那么锁将不会尝试在其析构函数中重新解锁。 我发现代码更符合我的口味,而不是将锁放入自己的块中。
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