Java 新手求解死锁要怎么处理

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这是一个创建于 1221 天前的主题，其中的信息可能已经有所发展或是发生改变。

题目是 10 个线程对应 20 个账号每个线程执行 100 次任意 2 个账号之间的金额交易

代码实现中遇到了死锁

假设线程 A 对账户 A 账户 B 执行交易（ A 转钱到 B ）线程 B 对账户 B 和账户 A 执行交易（ B 转钱到 A ）

我代码写的是先锁一个账号判断账号余额是否足够，然后锁另一个账户最后执行转账

但是同时有 2 个线程执行了 2 个相同账户的反向操作时，就会死锁了。

package cn.bobmao.logic.service;

import java.util.ArrayList;
import java.util.List;
import java.util.Random;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.atomic.AtomicInteger;

public class Test {
    class Account {
        int index;
        int balance = 100;

        public Account(int index) {
            this.index = index;
        }
    }

    public static void main(String[] args) throws InterruptedException {
        new Test().run();
    }

    public void run() throws InterruptedException {
        AtomicInteger sum = new AtomicInteger(0);
        List<Account> accounts = new ArrayList<>();
        for (int i = 0; i < 20; i++) {
            accounts.add(new Account(i));
        }
        ExecutorService executorService = Executors.newFixedThreadPool(10);

        CountDownLatch latch = new CountDownLatch(1000);
        for (int i = 0; i < 10; i++) {
            executorService.submit(new Pay(accounts, latch, sum));
        }

        latch.await();


        int count = 0;
        for (Account account : accounts) {
            count += account.balance;
        }

        System.out.println(count);


    }

    class Pay implements Runnable {
        List<Account> accounts;
        CountDownLatch latch;
        AtomicInteger sum;

        public Pay(List<Account> accounts, CountDownLatch latch, AtomicInteger sum) {
            this.accounts = accounts;
            this.latch = latch;
            this.sum = sum;
        }

        @Override
        public void run() {
            for (int i = 0; i < 100; i++) {
                int money = new Random().nextInt(100);
                // a->b
                int a = new Random().nextInt(20);
                int b = new Random().nextInt(20);
                latch.countDown();
                if (a == b) {
                    System.out.println("当前执行次数" + sum.getAndIncrement());

                    continue;
                }
                Account ac = accounts.get(a);
                synchronized (ac) {
                    System.out.println("锁 ac " + ac.index + " " + Thread.currentThread().getName());
                    if (ac.balance >= money) {
                        Account bc = accounts.get(b);
                        synchronized (bc) {
                            System.out.println("锁 bc " + ac.index + " " + Thread.currentThread().getName());
                            ac.balance = ac.balance - money;
                            bc.balance = bc.balance + money;
                            //latch.countDown();
                        }
                        System.out.println("解锁 bc " + ac.index + " " + Thread.currentThread().getName());
                    }
                }
                System.out.println("解锁 ac " + ac.index + " " + Thread.currentThread().getName());
                System.out.println("当前执行次数" + sum.getAndIncrement());
            }
        }
    }
}

9 条回复 • 2021-07-22 22:09:57 +08:00