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Semaphore底层源码实现原理解析

2020年09月28日  | 移动技术网IT编程  | 我要评论
一、SemaphoreDemoSemaphore的应用场景就是加锁、异步、控制并发数下面demo可以理解为抢夺共享资源import java.util.concurrent.Semaphore;public class SemaphoreDemo { public static void main(String[] args) { String[] name = {"小胖", "大花", "大肥", "小桌子", "小椅子"}; Semaphore sema

一、SemaphoreDemo

Semaphore的应用场景就是加锁异步控制并发数

下面demo可以理解为抢夺共享资源

import java.util.concurrent.Semaphore; public class SemaphoreDemo { public static void main(String[] args) { String[] name = {"小胖", "大花", "大肥", "小桌子", "小椅子"}; Semaphore semaphore = new Semaphore(3); for (int i = 0; i < 5; i++) { //五个人去争夺三个共享资源 //五个人一放学就去网吧抢机子,网吧只有三台电脑 new Thread(new Runnable() { @Override public void run() { try { System.out.println(Thread.currentThread().getName() + "进入网吧"); semaphore.acquire(); System.out.println(Thread.currentThread().getName() + "抢到了电脑"); Thread.sleep(5000); System.out.println(Thread.currentThread().getName() + "下机了"); } catch (InterruptedException e) { e.printStackTrace(); } finally { semaphore.release(); } } }, name[i]).start(); } } } 

运行结果

抢不到电脑的要等别人下机之后才能上网

在这里插入图片描述


二、加锁

在调用semaphore.acquire()方法后,进入InterruptedException(),然后点击进入tryAcquireShared()

public final void acquireSharedInterruptibly(int arg) throws InterruptedException { if (Thread.interrupted()) throw new InterruptedException(); if (tryAcquireShared(arg) < 0) doAcquireSharedInterruptibly(arg); } 

从tryAcquireShared()进入nonfairTryAcquireShared()

拿到state值后进行减一操作,然后CAS修改state值后返回

final int nonfairTryAcquireShared(int acquires) { for (;;) { int available = getState(); int remaining = available - acquires; if (remaining < 0 || compareAndSetState(available, remaining)) return remaining; } } 

如果state值小于0进入doAcquireSharedInterruptibly()

1)addWaiter(Node.SHARED)方法将当前线程加入同步队列,是一个双向链表,如果初始化为空,则进入enq();

2)enq()方法中:new Node()作为head,指向当前线程结点作为tail;

3)如果prev为head则进入tryAcquireShared(),然后尝试CAS修改state

4)将waitStatus设置为-1并且LockSupport.unpark()阻塞线程

private void doAcquireSharedInterruptibly(int arg) throws InterruptedException { final Node node = addWaiter(Node.SHARED); boolean failed = true; try { for (;;) { final Node p = node.predecessor(); // 1)如果prev为head则进入tryAcquireShared() if (p == head) { int r = tryAcquireShared(arg); // 3)尝试CAS修改state if (r >= 0) { setHeadAndPropagate(node, r); p.next = null; // help GC failed = false; return; } } if (shouldParkAfterFailedAcquire(p, node) && //4)将waitStatus设置为-1并且LockSupport.unpark()阻塞线程 parkAndCheckInterrupt()) throw new InterruptedException(); } } finally { if (failed) cancelAcquire(node); } } 

for(;)循环尝试修改state,成功后进入setHeadAndPropagate()

设置当前node为head,如果当前waitStauts为-1,则判断下一结点是否为null或者new Node(),如果是则直接唤醒

private void setHeadAndPropagate(Node node, int propagate) { Node h = head; // Record old head for check below setHead(node); /*
         * Try to signal next queued node if:
         *   Propagation was indicated by caller,
         *     or was recorded (as h.waitStatus either before
         *     or after setHead) by a previous operation
         *     (note: this uses sign-check of waitStatus because
         *      PROPAGATE status may transition to SIGNAL.)
         * and
         *   The next node is waiting in shared mode,
         *     or we don't know, because it appears null
         *
         * The conservatism in both of these checks may cause
         * unnecessary wake-ups, but only when there are multiple
         * racing acquires/releases, so most need signals now or soon
         * anyway.
         */ if (propagate > 0 || h == null || h.waitStatus < 0 || (h = head) == null || h.waitStatus < 0) { Node s = node.next; if (s == null || s.isShared()) doReleaseShared(); } } 

三、释放锁

进入releaseShared()

public final boolean releaseShared(int arg) { if (tryReleaseShared(arg)) { doReleaseShared(); return true; } return false; } 

tryReleaseShared()

释放锁,将state值加一,CAS修改成功后返回true

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; } } 

tryReleaseShared()返回true后进入doReleaseShared()

for(;;)循环判断head头结点不会null也不会尾结点,如果waitStatus为-1,则先CAS修改waitStatus为0,然后唤醒下一结点

private void doReleaseShared() { /*
         * Ensure that a release propagates, even if there are other
         * in-progress acquires/releases.  This proceeds in the usual
         * way of trying to unparkSuccessor of head if it needs
         * signal. But if it does not, status is set to PROPAGATE to
         * ensure that upon release, propagation continues.
         * Additionally, we must loop in case a new node is added
         * while we are doing this. Also, unlike other uses of
         * unparkSuccessor, we need to know if CAS to reset status
         * fails, if so rechecking.
         */ for (;;) { Node h = head; if (h != null && h != tail) { int ws = h.waitStatus; if (ws == Node.SIGNAL) { if (!compareAndSetWaitStatus(h, Node.SIGNAL, 0)) // CAS自旋修改waitStatus为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; } } 

如果CAS自旋修改waitStatus=0成功,进入unparkSuccessor()

保障判断waitStatus是否为0,判断下一结点是否为空,不为空则唤醒下一结点。

private void unparkSuccessor(Node node) { /*
         * If status is negative (i.e., possibly needing signal) try
         * to clear in anticipation of signalling.  It is OK if this
         * fails or if status is changed by waiting thread.
         */ int ws = node.waitStatus; if (ws < 0) compareAndSetWaitStatus(node, ws, 0); /*
         * Thread to unpark is held in successor, which is normally
         * just the next node.  But if cancelled or apparently null,
         * traverse backwards from tail to find the actual
         * non-cancelled successor.
         */ Node s = node.next; if (s == null || s.waitStatus > 0) { s = null; for (Node t = tail; t != null && t != node; t = t.prev) if (t.waitStatus <= 0) s = t; } if (s != null) LockSupport.unpark(s.thread); // 唤醒线程 } 

本文地址:https://blog.csdn.net/Small_Pang/article/details/108849209

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