目录
1:semphore的使用说明
参考这篇博客:https://www.cnblogs.com/crazymakercircle/p/13907012.html
2:semphore的使用案例
package com.saytoyou.com.thread;
import java.util.concurrent.Semaphore;
import java.util.concurrent.TimeUnit;
public class Student implements Runnable {
private String studentName;
private Semaphore semaphore;
/**
* 学生构造函数
*
* @param studentName
* @param semaphore
*/
public Student(String studentName, Semaphore semaphore) {
this.studentName = studentName;
this.semaphore = semaphore;
}
@Override
public void run() {
try {
semaphore.acquire();
System.out.println("我是"+studentName+"开始排队");
System.out.println("我是"+studentName + "开始打饭了" );
TimeUnit.SECONDS.sleep(3);
} catch (Exception e) {
e.printStackTrace();
} finally {
System.out.println("打饭完成" + studentName);
semaphore.release();
}
}
}
package com.saytoyou.com.thread;
import java.util.concurrent.Semaphore;
public class SemaphoreTest {
public static void main(String[] args) {
//模拟学生食堂打饭,设置三个窗口同时执行
Semaphore semaphore = new Semaphore(3,true);
for (int i = 0; i < 100; i++) {
new Thread(new Student("学生" + i, semaphore)).start();
}
}
}
3:semphore的源码分析
3.1:构造函数分析
public Semaphore(int permits, boolean fair) {
sync = fair ? new FairSync(permits) : new NonfairSync(permits);
}
构造semphore,permits是线程的个数,fair设置公平锁和非公平锁
3.2:acquire方法源码分析
private void doAcquireSharedInterruptibly(int arg)
throws InterruptedException {
final Node node = addWaiter(Node.SHARED);
boolean failed = true;
try {
for (;;) {
final Node p = node.predecessor();
if (p == head) {
int r = tryAcquireShared(arg);
if (r >= 0) {
setHeadAndPropagate(node, r);
p.next = null; // help GC
failed = false;
return;
}
}
if (shouldParkAfterFailedAcquire(p, node) &&
parkAndCheckInterrupt())
throw new InterruptedException();
}
} finally {
if (failed)
cancelAcquire(node);
}
}
分析:
1:首先将节点放入等待队列中,返回当前节点,注意第一次的头节点是new空节点,不是当前节点
2:拿到当前节点的前驱节点,判断前驱节点是否为头节点,如果为头节点则尝试去获取资源,获取成功当前state减一,返回剩余state,并尝试唤醒后续节点
如果不是头节点,则放入FIFO阻塞队列,等待后续唤醒
3:加入state的初始值为3,那么前三个线程都可以进入获取资源,后面的就需要等待线程释放,唤醒
4:第一次node节点进入,head是新节点,指向第一个节点,第二个节点跟这进来,状态改为signal,第三个节点同上,第一个节点进入后获取资源并唤醒第二个,第二个把头介点设置为自己,接着唤醒第三个,后面stagte资源不足了,就需要不断循环尝试获取了
3.3:tryAcquireShared源码分析
final int nonfairTryAcquireShared(int acquires) {
for (;;) {
int available = getState();
int remaining = available - acquires;
if (remaining < 0 ||
compareAndSetState(available, remaining))
return remaining;
}
}
分析:获取当前state的值,拿当前的state减去请求数,一般都是1,重新设置state的值,返回state的最新值
结合上面的分析为:进来之后进入阻塞队列中,判断节点的前驱节点是不是head如果是,就尝试获取资源,获取成功之后,唤醒后续节点
流程图如下:

4:释放资源源码分析
protected final boolean tryReleaseShared(int releases) {
for (;;) {
int current = getState();
int next = current + releases;
if (next < current) // overflow
throw new Error("Maximum permit count exceeded");
if (compareAndSetState(current, next))
return true;
}
}
释放资源,state的值加一,这样后续第四个线程被唤醒,其实是第一个线程释放state,然后唤醒第四个线程的
4.1 唤醒后续的等待线程
private void doReleaseShared() {
for (;;) {
Node h = head;
if (h != null && h != tail) {
int ws = h.waitStatus;
if (ws == Node.SIGNAL) {
if (!compareAndSetWaitStatus(h, Node.SIGNAL, 0))
continue; // loop to recheck cases
unparkSuccessor(h);
}
else if (ws == 0 &&
!compareAndSetWaitStatus(h, 0, Node.PROPAGATE))
continue; // loop on failed CAS
}
if (h == head) // loop if head changed
break;
}
}
此时head是指向第三个线程的,状态为signal,唤醒后一个线程
原创文章,作者:3628473679,如若转载,请注明出处:https://blog.ytso.com/tech/pnotes/244276.html