semaphore

I want some clarification regarding mutex and semaphore. My question is, What mutex actually do when a thread tries to enter a region locked by a mutex, a. it waits for the lock to be released? or b. it goes to sleep until the lock is released. In that case how it is wake up again when the lock is released? Same question as 1, but in this case it is semaphore. Can you give me some code regarding busy waiting in pthread in C, and also a case where thread goes to sleep instead of waiting? does sleep mean it is blocked or sleeping is another kind of busy waiting? i want to know some programs

I wanted to know what would be better/faster to use POSIX calls like pthread_once() and sem_wait() or the dispatch_* functions, so I created a little test and am surprised at the results (questions and results are at the end). In the test code I am using mach_absolute_time() to time the calls. I really don’t care that this is not exactly matching up with nano-seconds; I am comparing the values with each other so the exact time units don't matter, only the differences between the interval do. The numbers in the results section are repeatable and not averaged; I could have averaged the times but

i have two methods -(void) a { @synchronized(self) { // critical section 1 } } -(void) b { @synchronized(self) { // critical section 2 } } now my question is if a thread is in critical section 1. will the critical section 2 be locked for other threads or other threads can access critical section 2. Critical section 2 will be blocked to other threads, as well, since you're synchronizing on the same object ( self ). 来源： https://stackoverflow.com/questions/2810459/objective-c-synchronized-how-does-it-work

I am writing a program (for homework) that simulates a unisex bathroom. Only 4 people are allowed at a time and men and woman cannot enter if the other sex is already using the bathroom. My problem is with allowing a max of 4 people in the bathroom. As you can see from the output, only 1 person is getting into the restroom at a time. Here is my code: const int Delayx = 60; int i; int restroom = 0; int Menwaiting = 0; int Womenwaiting = 0; semaphore max_capacity; semaphore woman; semaphore man; semaphore mutex; semaphore restroomcount; void Delay(void) { int DelayTime; DelayTime = random(Delayx

信号量Semaphore 同进程的一样 Semaphore管理一个内置的计数器， 每当调用acquire()时内置计数器-1； 调用release() 时内置计数器+1； 计数器不能小于0；当计数器为0时，acquire()将阻塞线程直到其他线程调用release()。 from threading import Thread,currentThread,Semaphore import time def task(): sm.acquire() print(f'{currentThread().name}在执行') time.sleep(3) sm.release() sm=Semaphore(5) for i in range(15): t=Thread(target=task) t.start() 来源： https://www.cnblogs.com/aden668/p/11542639.html

java5 中 ，提供了几个并发工具类 ，Semaphore CountDownLatch CyclicBarrier，在并发编程中非常实用。前两者通过 内部类sync 继承AQS，使用共享资源的模式，AQS的实现可参考我的另一篇 AQS 实现分析 ，前两者根据各自功能需求 ， 各自内部实现tryAcquireShared（获取资源）、tryReleaseShared（释放）。来定义什么条件 下来获取与释放。而CyclicBarrier内部通过Reentrantlock与Condition组合的方式实现。 Semaphore 与Reentrantlock类似，也有公平和非公平的机制。这里就不在分析了，默认是非公平的。 通过acquire 获取 Semphore public void acquire() throws InterruptedException { sync.acquireSharedInterruptibly(1); //非独占模式 } AQS public final void acquireSharedInterruptibly(int arg) throws InterruptedException { if (Thread.interrupted()) throw new InterruptedException(); if

Here is the code I have but I don't understand what SemaphoreSlim is doing. async Task WorkerMainAsync() { SemaphoreSlim ss = new SemaphoreSlim(10); List<Task> trackedTasks = new List<Task>(); while (DoMore()) { await ss.WaitAsync(); trackedTasks.Add(Task.Run(() => { DoPollingThenWorkAsync(); ss.Release(); })); } await Task.WhenAll(trackedTasks); } void DoPollingThenWorkAsync() { var msg = Poll(); if (msg != null) { Thread.Sleep(2000); // process the long running CPU-bound job } } What does await ss.WaitAsync(); & ss.Release(); do? I guess that if I run 50 threads at a time then write code