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Java并发编程 - 线程不安全类与解决方案（List）

三维控件研究 2022-05-18 阅读 60
Java并发编程List线程不安全类与解决方案并发多线程Java编程语言


List

package com.mmall.concurrency.example.commonUnsafe;

import com.mmall.concurrency.annoations.NotThreadSafe;
import lombok.extern.slf4j.Slf4j;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;

@Slf4j
@NotThreadSafe
public class ArrayListExample {

    // 请求总数
    public static int clientTotal = 5000;

    // 同时并发执行的线程数
    public static int threadTotal = 200;

    private static List<Integer> list = new ArrayList<>();

    public static void main(String[] args) throws Exception {
        ExecutorService executorService = Executors.newCachedThreadPool();
        final Semaphore semaphore = new Semaphore(threadTotal);
        final CountDownLatch countDownLatch = new CountDownLatch(clientTotal);
        for (int i = 0; i < clientTotal; i++) {
            final int count = i;
            executorService.execute(() -> {
                try {
                    semaphore.acquire();
                    update(count);
                    semaphore.release();
                } catch (Exception e) {
                    log.error("exception", e);
                }
                countDownLatch.countDown();
            });
        }
        countDownLatch.await();
        executorService.shutdown();
        log.info("size:{}", list.size());
    }

    private static void update(int i) {
        list.add(i);
    }
}

// 输出
size:4988


解决方案一（同步容器：Vector）

package com.mmall.concurrency.example.syncContainer;

import com.mmall.concurrency.annoations.ThreadSafe;
import lombok.extern.slf4j.Slf4j;
import java.util.List;
import java.util.Vector;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;

@Slf4j
@ThreadSafe
public class VectorExample1 {

    // 请求总数
    public static int clientTotal = 5000;

    // 同时并发执行的线程数
    public static int threadTotal = 200;

    private static List<Integer> list = new Vector<>();

    public static void main(String[] args) throws Exception {
        ExecutorService executorService = Executors.newCachedThreadPool();
        final Semaphore semaphore = new Semaphore(threadTotal);
        final CountDownLatch countDownLatch = new CountDownLatch(clientTotal);
        for (int i = 0; i < clientTotal; i++) {
            final int count = i;
            executorService.execute(() -> {
                try {
                    semaphore.acquire();
                    update(count);
                    semaphore.release();
                } catch (Exception e) {
                    log.error("exception", e);
                }
                countDownLatch.countDown();
            });
        }
        countDownLatch.await();
        executorService.shutdown();
        log.info("size:{}", list.size());
    }

    private static void update(int i) {
        list.add(i);
    }
}

// 输出
size:5000


解决方案二（同步容器：synchronizedList）

package com.mmall.concurrency.example.syncContainer;

import com.google.common.collect.Lists;
import com.mmall.concurrency.annoations.ThreadSafe;
import lombok.extern.slf4j.Slf4j;
import java.util.Collections;
import java.util.List;
import java.util.Vector;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;

@Slf4j
@ThreadSafe
public class CollectionsExample1 {

    // 请求总数
    public static int clientTotal = 5000;

    // 同时并发执行的线程数
    public static int threadTotal = 200;

    private static List<Integer> list = Collections.synchronizedList(Lists.newArrayList());

    public static void main(String[] args) throws Exception {
        ExecutorService executorService = Executors.newCachedThreadPool();
        final Semaphore semaphore = new Semaphore(threadTotal);
        final CountDownLatch countDownLatch = new CountDownLatch(clientTotal);
        for (int i = 0; i < clientTotal; i++) {
            final int count = i;
            executorService.execute(() -> {
                try {
                    semaphore.acquire();
                    update(count);
                    semaphore.release();
                } catch (Exception e) {
                    log.error("exception", e);
                }
                countDownLatch.countDown();
            });
        }
        countDownLatch.await();
        executorService.shutdown();
        log.info("size:{}", list.size());
    }

    private static void update(int i) {
        list.add(i);
    }
}

// 输出
szie:5000


解决方案三（并发容器：CopyOnWriteArrayList）

package com.mmall.concurrency.example.concurrent;

import com.mmall.concurrency.annoations.ThreadSafe;
import lombok.extern.slf4j.Slf4j;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.CopyOnWriteArrayList;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;

@Slf4j
@ThreadSafe
public class CopyOnWriteArrayListExample {

    // 请求总数
    public static int clientTotal = 5000;

    // 同时并发执行的线程数
    public static int threadTotal = 200;

    private static List<Integer> list = new CopyOnWriteArrayList<>();

    public static void main(String[] args) throws Exception {
        ExecutorService executorService = Executors.newCachedThreadPool();
        final Semaphore semaphore = new Semaphore(threadTotal);
        final CountDownLatch countDownLatch = new CountDownLatch(clientTotal);
        for (int i = 0; i < clientTotal; i++) {
            final int count = i;
            executorService.execute(() -> {
                try {
                    semaphore.acquire();
                    update(count);
                    semaphore.release();
                } catch (Exception e) {
                    log.error("exception", e);
                }
                countDownLatch.countDown();
            });
        }
        countDownLatch.await();
        executorService.shutdown();
        log.info("size:{}", list.size());
    }

    private static void update(int i) {
        list.add(i);
    }
}

// 输出
size:5000

分析

  • CopyOnWriteArrayList 写操作时复制，当有新元素添加到集合中时，从原有的数组中拷贝一份出来，然后在新的数组上作写操作，然后将原来的数组指向新的数组。整个数组的add操作都是在锁的保护下进行的，防止并发时复制多份副本。读操作是在原数组中进行，不需要加锁。

缺点

  1. 写操作时复制消耗内存
  2. 不能用于实时读的场景
  3. 由于复制和add操作等需要时间，故读取时可能读到旧值。
  4. 能做到最终一致性，但无法满足实时性的要求，更适合读多写少的场景。
  5. 如果无法知道数组有多大，或者add、set操作有多少，慎用此类。

设计思想

  1. 读写分离
  2. 最终一致性
  3. 使用时另外开辟空间，防止并发冲突


附1：Vector 线程不安全情景

package com.mmall.concurrency.example.syncContainer;

import com.mmall.concurrency.annoations.NotThreadSafe;
import java.util.Vector;

@NotThreadSafe
public class VectorExample2 {

    private static Vector<Integer> vector = new Vector<>();

    public static void main(String[] args) {

        while (true) {

            for (int i = 0; i < 10; i++) {
                vector.add(i);
            }

            Thread thread1 = new Thread() {
                public void run() {
                    for (int i = 0; i < vector.size(); i++) {
                        vector.remove(i);
                    }
                }
            };

            Thread thread2 = new Thread() {
                public void run() {
                    for (int i = 0; i < vector.size(); i++) {
                        vector.get(i);
                    }
                }
            };
            thread1.start();
            thread2.start();
        }
    }
}

Java并发编程 - 线程不安全类与解决方案（List）_多线程

  • 输出结果报错，因为 remove 和 get 分别是不同对象调用的不同 sync 方法，很有可能出现 remove 在 get 之前操作。


附2：Vector for & foerach & iterator

package com.mmall.concurrency.example.syncContainer;

import java.util.Iterator;
import java.util.Vector;

public class VectorExample3 {

    // java.util.ConcurrentModificationException
    private static void test1(Vector<Integer> v1) { // foreach
        for(Integer i : v1) {
            if (i.equals(3)) {
                v1.remove(i);
            }
        }
    }

    // java.util.ConcurrentModificationException
    private static void test2(Vector<Integer> v1) { // iterator
        Iterator<Integer> iterator = v1.iterator();
        while (iterator.hasNext()) {
            Integer i = iterator.next();
            if (i.equals(3)) {
                v1.remove(i);
            }
        }
    }

    // success
    private static void test3(Vector<Integer> v1) { // for
        for (int i = 0; i < v1.size(); i++) {
            if (v1.get(i).equals(3)) {
                v1.remove(i);
            }
        }
    }

    public static void main(String[] args) {

        Vector<Integer> vector = new Vector<>();
        vector.add(1);
        vector.add(2);
        vector.add(3);
        test1(vector);
    }
}

Java并发编程 - 线程不安全类与解决方案（List）_Java并发编程_02

  • 输出结果

Java并发编程 - 线程不安全类与解决方案（List）_多线程_03

  • 源码分析，因为 modCount 只要更新操作会被修改导致这里不相等抛出异常。

结论

使用foreach，iterator以及for循环对集合类遍历的同时进行修改

  1. 在foreach、iterator迭代器循环集合的时候，在遍历过程中尽量不要做更新操作。
  2. 如果一定要做的话，在遍历过程中，只做标记，遍历完成后再更新。
  3. 遍历过程中(foeach,iterator)更新会导致ConcurrentModificationException
  4. 可以用for循环做遍历过程中的增删操作


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