java 多线程知识梳理4
常用并发辅助类 CountDownLatch Semaphore CyclicBarrier, 都基于ReentrantLock实现。
1 Semaphores are often used to restrict the number of threads than can
* access some (physical or logical) resource. For example, here is
* a class that uses a semaphore to control access to a pool of items:
Semaphores 信号量, 相对常规同步块只能有一个线程对资源进行操作,Semaphores能允许多个线程同时操作资源。
public Semaphore(int permits) { //参数permits表示许可数目,即同时可以允许多少线程进行访问 sync = new NonfairSync(permits);}public Semaphore(int permits, boolean fair) { //这个多了一个参数fair表示是否是公平的,即等待时间越久的越先获取许可 sync = (fair)? new FairSync(permits) : new NonfairSync(permits);}public void acquire() throws InterruptedException { } //获取一个许可public void acquire(int permits) throws InterruptedException { } //获取permits个许可public void release() { } //释放一个许可public void release(int permits) { } //释放permits个许可public boolean tryAcquire() { }; //尝试获取一个许可,若获取成功,则立即返回true,若获取失败,则立即返回falsepublic boolean tryAcquire(long timeout, TimeUnit unit) throws InterruptedException { }; //尝试获取一个许可,若在指定的时间内获取成功,则立即返回true,否则则立即返回falsepublic boolean tryAcquire(int permits) { }; //尝试获取permits个许可,若获取成功,则立即返回true,若获取失败,则立即返回falsepublic boolean tryAcquire(int permits, long timeout, TimeUnit unit) throws InterruptedException { }; //尝试获取permits个许可,若在指定的时间内获取成功,则立即返回true,否则则立即返回false * class Pool {
* private static final int MAX_AVAILABLE = 100;
* private final Semaphore available = new Semaphore(MAX_AVAILABLE, true);
*
* public Object getItem() throws InterruptedException {
* available.acquire();//允许前100个线程进入方法,之后的线程阻塞
* return getNextAvailableItem();
* }
*
* public void putItem(Object x) {
* if (markAsUnused(x))
* available.release();//是否一个许可,之后阻塞的线程可以调用getItem方法
* }
*
* // Not a particularly efficient data structure; just for demo
*
* protected Object[] items = ... whatever kinds of items being managed
* protected boolean[] used = new boolean[MAX_AVAILABLE];
*
* protected synchronized Object getNextAvailableItem() {
* for (int i = 0; i < MAX_AVAILABLE; ++i) {
* if (!used[i]) {
* used[i] = true;
* return items[i];
* }
* }
* return null; // not reached
* }
*
* protected synchronized boolean markAsUnused(Object item) {
* for (int i = 0; i < MAX_AVAILABLE; ++i) {
* if (item == items[i]) {
* if (used[i]) {
* used[i] = false;
* return true;
* } else
* return false;
* }
* }
* return false;
* }
*
* }
* private static final int MAX_AVAILABLE = 100;
* private final Semaphore available = new Semaphore(MAX_AVAILABLE, true);
*
* public Object getItem() throws InterruptedException {
* available.acquire();//允许前100个线程进入方法,之后的线程阻塞
* return getNextAvailableItem();
* }
*
* public void putItem(Object x) {
* if (markAsUnused(x))
* available.release();//是否一个许可,之后阻塞的线程可以调用getItem方法
* }
*
* // Not a particularly efficient data structure; just for demo
*
* protected Object[] items = ... whatever kinds of items being managed
* protected boolean[] used = new boolean[MAX_AVAILABLE];
*
* protected synchronized Object getNextAvailableItem() {
* for (int i = 0; i < MAX_AVAILABLE; ++i) {
* if (!used[i]) {
* used[i] = true;
* return items[i];
* }
* }
* return null; // not reached
* }
*
* protected synchronized boolean markAsUnused(Object item) {
* for (int i = 0; i < MAX_AVAILABLE; ++i) {
* if (item == items[i]) {
* if (used[i]) {
* used[i] = false;
* return true;
* } else
* return false;
* }
* }
* return false;
* }
*
* }
2 CountDownLatch 利用它可以实现类似计数器的功能
public CountDownLatch(int count) { }; //参数count为计数值public void await() throws InterruptedException { }; //调用await()方法的线程会被挂起,它会等待直到count值为0才继续执行public boolean await(long timeout, TimeUnit unit) throws InterruptedException { }; //和await()类似,只不过等待一定的时间后count值还没变为0的话就会继续执行public void countDown() { }; //将count值减1public class Test { public static void main(String[] args) { final CountDownLatch latch = new CountDownLatch(2); new Thread(){ public void run() { try { System.out.println("子线程"+Thread.currentThread().getName()+"正在执行"); Thread.sleep(3000); System.out.println("子线程"+Thread.currentThread().getName()+"执行完毕"); latch.countDown(); } catch (InterruptedException e) { e.printStackTrace(); } }; }.start(); new Thread(){ public void run() { try { System.out.println("子线程"+Thread.currentThread().getName()+"正在执行"); Thread.sleep(3000); System.out.println("子线程"+Thread.currentThread().getName()+"执行完毕"); latch.countDown(); } catch (InterruptedException e) { e.printStackTrace(); } }; }.start(); try { System.out.println("等待2个子线程执行完毕..."); latch.await(); System.out.println("2个子线程已经执行完毕"); System.out.println("继续执行主线程"); } catch (InterruptedException e) { e.printStackTrace(); } } }3 CyclicBarrier 栅栏,通过它可以实现让一组线程等待至某个状态之后再全部同时执行。 CyclicBarrier可以重用,CountDownLatch不可以
public CyclicBarrier(int parties, Runnable barrierAction) {//参数parties指让多少个线程或者任务等待至barrier状态;参数barrierAction为当这些线程都达到barrier状态时会执行的内容。}public CyclicBarrier(int parties) {}public int await() throws InterruptedException, BrokenBarrierException { };//用来挂起当前线程,直至所有线程都到达barrier状态再同时执行后续任务;public int await(long timeout, TimeUnit unit)throws InterruptedException,BrokenBarrierException,TimeoutException { }; * class Solver {
* final int N;
* final float[][] data;
* final CyclicBarrier barrier;
*
* class Worker implements Runnable {
* int myRow;
* Worker(int row) { myRow = row; }
* public void run() {
* while (!done()) {
* processRow(myRow);
*
* try {
* barrier.await();//设置当前线程为barried状态,barrier内置的计数器+1,直到N
* } catch (InterruptedException ex) {
* return;
* } catch (BrokenBarrierException ex) {
* return;
* }
* }
* }
* }
*
* public Solver(float[][] matrix) {
* data = matrix;
* N = matrix.length;
* barrier = new CyclicBarrier(N,
* new Runnable() {//当N个线程处于barried状态后,执行后续的mergeRows任务。
* public void run() {
* mergeRows(...);
* }
* });
* for (int i = 0; i < N; ++i)
* new Thread(new Worker(i)).start();
* final int N;
* final float[][] data;
* final CyclicBarrier barrier;
*
* class Worker implements Runnable {
* int myRow;
* Worker(int row) { myRow = row; }
* public void run() {
* while (!done()) {
* processRow(myRow);
*
* try {
* barrier.await();//设置当前线程为barried状态,barrier内置的计数器+1,直到N
* } catch (InterruptedException ex) {
* return;
* } catch (BrokenBarrierException ex) {
* return;
* }
* }
* }
* }
*
* public Solver(float[][] matrix) {
* data = matrix;
* N = matrix.length;
* barrier = new CyclicBarrier(N,
* new Runnable() {//当N个线程处于barried状态后,执行后续的mergeRows任务。
* public void run() {
* mergeRows(...);
* }
* });
* for (int i = 0; i < N; ++i)
* new Thread(new Worker(i)).start();
system.out.println("cyclicBarrier重复使用");
* for (int i = 0; i < N; ++i)
* new Thread(new Worker(i)).start();
*
* waitUntilDone();
* }
* }
* new Thread(new Worker(i)).start();
*
* waitUntilDone();
* }
* }
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