我们何时应该使用互斥锁?什么时候应该使用信号量?
在java中是否存在Mutex对象或创建一个?我问,因为用1许可证初始化的信号量对象对我没有帮助.想想这个案例:
try {
semaphore.acquire();
//do stuff
semaphore.release();
} catch (Exception e) {
semaphore.release();
}
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如果在第一次获取时发生异常,则catch块中的释放将增加许可,并且信号量不再是二进制信号量.
请问是正确的方法吗?
try {
semaphore.acquire();
//do stuff
} catch (Exception e) {
//exception stuff
} finally {
semaphore.release();
}
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上面的代码会确保信号量是二进制的吗?
java mutex semaphore java.util.concurrent concurrent-programming
我正在学习Java中的Mutex和Semaphore.所以我想弄脏手.
我理解互斥锁是信号量,具有来自此链接的单一许可,而互斥锁具有来自此链接的所有权概念.
为了证明所有权,我写了下面的程序,并在下面输出.当我发布的数量超过收购数量时,它实际上会增加许可证的数量.
以下是该计划.
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.concurrent.Semaphore;
public class MutexOwnerShipDemo {
public static void main(String[] args) {
Semaphore mutex = new Semaphore(1);
final ExecutorService es = Executors.newSingleThreadExecutor();
try {
// acquire mutex lock permit in main thread
mutex.acquire();
// release the lock in different thread
Future mutexWait = es.submit(() -> mutex.release());
mutexWait.get();
System.out.println("mutexWait.isDone() " + mutexWait.isDone() );
System.out.println("successfully released permits for mutex \n " +
"available permits are " + mutex.availablePermits()); …Run Code Online (Sandbox Code Playgroud)