mutex

问题 This is an interview question . In general the deadlock between 2 threads is generated when thread1 locks mutex1,and a moment before it tries to lock mutex2 ,thread 2 locks mutex2.After that tread 2 wants to lock mutex1.So they wait for each other forever. The question was "Can you gave a scenario of deadlock with one mutex and any number of threads?" 回答1: It depends on how you define "deadlock" I guess, but I could see one possibility: Thread A grabs the mutex Thread B waits for mutex Thread

问题 This is an interview question . In general the deadlock between 2 threads is generated when thread1 locks mutex1,and a moment before it tries to lock mutex2 ,thread 2 locks mutex2.After that tread 2 wants to lock mutex1.So they wait for each other forever. The question was "Can you gave a scenario of deadlock with one mutex and any number of threads?" 回答1: It depends on how you define "deadlock" I guess, but I could see one possibility: Thread A grabs the mutex Thread B waits for mutex Thread

问题 I have a problem when i want use mutex to force single instance of my program. I a winform App with a WebBrowser Control. I need to autorestart if certain conditions are met. My Problem is that if i restart the application the mutex want let me open a new instance (this doesn't always happen) maybe cause browser has some async method like navigate. But i release the Mutex in Exit Application Event. This is my Code: static Mutex _mutex = new Mutex (false, "myprogramkey"); [STAThread] private

问题 I have a problem when i want use mutex to force single instance of my program. I a winform App with a WebBrowser Control. I need to autorestart if certain conditions are met. My Problem is that if i restart the application the mutex want let me open a new instance (this doesn't always happen) maybe cause browser has some async method like navigate. But i release the Mutex in Exit Application Event. This is my Code: static Mutex _mutex = new Mutex (false, "myprogramkey"); [STAThread] private

问题 I created one thread in my main program, thread execution has to stop once the main program will terminate. I am using reader.join(); to terminate the thread execution. But it is not stopping the execution. I tried with below-mentioned code, I am using thread.join(); function, but it is failed to terminate a thread. And after the main program also my thread is kept executing. #include <algorithm> #include <array> #include <atomic> #include <mutex> #include <queue> #include <cstdint> #include

问题 I created one thread in my main program, thread execution has to stop once the main program will terminate. I am using reader.join(); to terminate the thread execution. But it is not stopping the execution. I tried with below-mentioned code, I am using thread.join(); function, but it is failed to terminate a thread. And after the main program also my thread is kept executing. #include <algorithm> #include <array> #include <atomic> #include <mutex> #include <queue> #include <cstdint> #include

问题 There is a shared_mutex class planned for C++17. And shared_timed_mutex already in C++14. (Who knows why they came in that order, but whatever.) Then there is a recursive_mutex and a recursive_timed_mutex since C++11. What I need is a shared_recursive_mutex . Did I miss something in the standard or do I have to wait another three years for a standardized version of that? If there is currently no such facility, what would be a simple (first priority) and efficient (2nd priority) implementation

问题 Even if the shared variable is atomic, it must be modified under the mutex in order to correctly publish the modification to the waiting thread. Any thread that intends to wait on std::condition_variable has to acquire a std::unique_lock, on the same mutex as used to protect the shared variable http://en.cppreference.com/w/cpp/thread/condition_variable I understand that by protecting the std::condition_variable with a mutex we are protected against missing a notify if the waiting thread is

问题 I'm trying to solve the thread-ring problem. In each thread I read the token value if it is not mine, check if it's the end of the program if it is then finish the thread otherwise, read again and repeat if it is mine (i.e. has my id) then acquire the write lock, increase the value of the token, check if it's the end then tell main thread that I finished it and finish the current thread loop If it not over, then release the write lock, and start to read again There is no unlock. Is there any

问题 I'm trying to solve the thread-ring problem. In each thread I read the token value if it is not mine, check if it's the end of the program if it is then finish the thread otherwise, read again and repeat if it is mine (i.e. has my id) then acquire the write lock, increase the value of the token, check if it's the end then tell main thread that I finished it and finish the current thread loop If it not over, then release the write lock, and start to read again There is no unlock. Is there any