semaphore

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Is it true that C++0x will come without semaphores? There are already some questions on Stack Overflow regarding the use of semaphores. I use them (posix semaphores) all the time to let a thread wait for some event in another thread: void thread0(...) { doSomething0(); event1.wait(); ... } void thread1(...) { doSomething1(); event1.post(); ... } If I would do that with a mutex: void thread0(...) { doSomething0(); event1.lock(); event1.unlock(); ... } void thread1(...) { event1.lock(); doSomethingth1(); event1.unlock(); ... } Problem: It's

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 由 翻译 强力驱动 问题: The documentation for the .NET Semaphore class states that: There is no guaranteed order, such as FIFO or LIFO, in which blocked threads enter the semaphore. In this case, if I want a guaranteed order (either FIFO or LIFO), what are my options? Is this something that just isn't easily possible? Would I have to write my own Semaphore? I assume that would be pretty advanced? Thanks, Steve 回答1: See this : The FifoSemaphore works exactly like a normal Semaphore but also guarantees that tokens are served out to acquirers in the order

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am trying to create a method which makes use of UIView's "+animateWithDuration:animations:completion" method to perform animations, and wait for completion. I am well aware that I could just place the code that would normally come after it in a completion block, but I would like to avoid this because there is a substantial amount of code after it including more animations, which would leave me with nested blocks. I tried to implement this method as below using a semaphore, but I don't think this is the best way to do it, especially because

1、使用信号量Semaphore来实现有界缓存 2、代码如下： /** * BoundedBuffer * <p/> * Bounded buffer using \Semaphore * * @author Brian Goetz and Tim Peierls */ @ThreadSafe public class SemaphoreBoundedBuffer < E > { private final Semaphore availableItems , availableSpaces ; @GuardedBy ( "this" ) private final E [] items ; @GuardedBy ( "this" ) private int putPosition = 0 , takePosition = 0 ; public SemaphoreBoundedBuffer ( int capacity ) { if ( capacity <= 0 ) throw new IllegalArgumentException () ; availableItems = new Semaphore ( 0 ) ; availableSpaces = new Semaphore ( capacity ) ; items = ( E []) new Object [

今天业务需求开发需要开发一个洗数据的小功能，大致业务是有百万级别的数据需要清洗，需要开发一个小功能，循环遍历百万数据调用一个服务接口清晰数据。考虑到接口的并发量，访问量不能太大， 整了一两个小时做了一个并发控制的小程序。 public static class CleanTask implements Runnable { private static Semaphore semaphore = new Semaphore( 5 , true ) ; private Runnable runnable ; private CountDownLatch latch ; public CleanTask (CountDownLatch latch , Runnable runnable) { this . runnable = runnable ; this . latch = latch ; } @Override public void run () { try { semaphore .acquire() ; runnable .run() ; semaphore .release() ; } catch (InterruptedException ex) { LOG .error( "线程中断异常" , ex) ; } finally { latch .countDown()

Semaphore是计数信号量，它管理了一定数量的许可证，每个线程要使用资源必须获得许可证，使用完了再释放许可证。当许可证被发放完了时，其他线程再想获得，必须等待，即进入了阻塞状态，当有线程释放了许可证时，它才能获得执行。 方法摘要 acquire () 中断 。 中断 。 acquireUninterruptibly () availablePermits () drainPermits () getQueuedThreads () getQueueLength () hasQueuedThreads () isFair () true 。 release () toString () tryAcquire () 中断 ，则从此信号量获取给定数目的许可。 中断 ，则从此信号量获取一个许可。 比如某家餐厅只有五个餐桌，来了一百个人，每个人都想独占一张餐桌，这是就必须要等待有人吃完释放了餐桌，其他人才能吃，每次也只能有五个人同时吃，因为只有五张餐桌。 public class SemaphoreTest { public static void main(String[] args) { Semaphore semaphore = new Semaphore(5); for (int i = 0; i < 100; i++) { new Thread(new Runnable() {

Condition ConditionWait public class ConditionDemoWait implements Runnable{ 　　private Lock lock;　　private Condition condition; 　　public ConditionDemoWait(Lock lock, 　　Condition condition){ 　　　　this.lock=lock; 　　　　this.condition=condition; 　　} 　　@Override 　　public void run() { 　　　　System.out.println("begin -ConditionDemoWait"); 　　　　ry { 　　　　　　lock.lock(); 　　　　　　condition.await(); 　　　　　　System.out.println("end -ConditionDemoWait"); 　　　　} catch (InterruptedException e) { 　　　　　　e.printStackTrace(); 　　　　}finally { 　　　　　　lock.unlock(); 　　　　}　　} } ConditionSignal public class ConditionDemoSignal

Semaphore: 信号量 Semaphore: 可以指定多个线程同时访问某一资源。 一）、构造方法 //int permits:线程的准入数，即一个资源同时可以允许多少个线程访问 Semaphore semaphore = new Semaphore(int permits); //boolean fair，指明锁的规则， false: 非公平锁， true: 公平锁 Semaphore semaphore = new Semaphore(int permits, boolean fair) 公平锁： 锁的顺序与线程的执行顺序有关 非公平锁：锁的执行顺序与线程的执行顺序无关 默认使用非公平锁 。 二）、获取许可的方法 1)、acquire()：获取一个许可，如果获取失败，则线程等待，等待期间可以响应中 断。 public void acquire() throws InterruptedException { //1.获取共享可中断的 sync.acquireSharedInterruptibly(1); } public final void acquireSharedInterruptibly(int arg) throws InterruptedException { //2.优先判断线程的中断状态 if (Thread.interrupted()) throw new

public class Semaphore implements java.io.Serializable { private static final long serialVersionUID = -3222578661600680210L; /** All mechanics via AbstractQueuedSynchronizer subclass */ private final Sync sync; /** * 继承AQS以实现信号量机制，使用AQS的state属性代表允许使用的信号量的数量，该Sync类有两个 * 派生类，其中一个是公平模式的，另一个是非公平模式的 */ abstract static class Sync extends AbstractQueuedSynchronizer { private static final long serialVersionUID = 1192457210091910933L; Sync(int permits) { setState(permits); } final int getPermits() { return getState(); } //非公平模式的获取信号量 final int nonfairTryAcquireShared(int acquires) { for (;;) { int

Can the semaphore be lower than 0? I mean, say I have a semaphore with N=3 and I call "down" 4 times, then N will remain 0 but one process will be blocked? And same the other way, if in the beginning I call up, can N be higher than 3? Because as I see it, if N can be higher than 3 if in the beginning I call up couple of times, then later on I could call down more times than I can, thus putting more processes in the critical section then the semaphore allows me. If someone would clarify it a bit for me I will much appreciate. Greg Calling down when it's 0 should not work. Calling up when it's 3