我正在閱讀使用condition_variable
here的示例代碼。我張貼下面的代碼:爲什麼lock_guard可以通過unique_lock獲取已經鎖定的互斥鎖?
std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;
void worker_thread()
{
// Wait until main() sends data
std::cout << "------------------------\n";
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return ready;});
// after the wait, we own the lock.
std::cout << "Worker thread is processing data\n";
data += " after processing";
// Send data back to main()
processed = true;
std::cout << "Worker thread signals data processing completed\n";
// Manual unlocking is done before notifying, to avoid waking up
// the waiting thread only to block again (see notify_one for details)
lk.unlock();
cv.notify_one();
}
int main()
{
std::thread worker(worker_thread);
data = "Example data";
// send data to the worker thread
{
std::lock_guard<std::mutex> lk(m);
ready = true;
std::cout << "main() signals data ready for processing\n";
}
cv.notify_one();
// wait for the worker
{
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return processed;});
}
std::cout << "Back in main(), data = " << data << '\n';
worker.join();
return 0;
}
我的問題是worker_thread
是首次推出,所以我將承擔互斥m
由worker_thread
鎖定,但爲什麼在main
互斥m
仍然可以通過lock_guard
鎖定?
是不是*互斥*的整點多線程可以嘗試將其鎖定? –
如果我有意讓'main'睡眠一段時間以確保'worker_thread'首先運行,它仍然輸出'main()信號數據準備好處理',爲什麼? – Allanqunzi