21

我想知道在以下示例中避免死锁的替代方法是什么。下面的例子是一个典型的银行账户转账死锁问题。在实践中解决它的一些更好的方法是什么?

class Account {
     double balance;
     int id;
     public Account(int id, double balance){
          this.balance = balance;
          this.id = id;
     }
     void withdraw(double amount){
          balance -= amount;
     } 
     void deposit(double amount){
          balance += amount;
     }
}
class Main{
     public static void main(String [] args){
           final Account a = new Account(1,1000);
           final Account b = new Account(2,300);
           Thread a = new Thread(){
                 public void run(){
                     transfer(a,b,200);
                 }
           };
           Thread b = new Thread(){
                 public void run(){
                     transfer(b,a,300);
                 }
           };
           a.start();
           b.start();
     }
     public static void transfer(Account from, Account to, double amount){
          synchronized(from){
               synchronized(to){
                    from.withdraw(amount);
                    to.deposit(amount);
               }
          }
     }
}
  

我想知道如果我在我的传输方法中将嵌套锁分开,是否会解决死锁问题,如下所示

 synchronized(from){
      from.withdraw(amount);
 }
 synchronized(to){
      to.deposit(amount);
 }
4

6 回答 6

35

对帐户进行排序。死锁来自帐户的顺序(a,b vs b,a)。

所以试试:

 public static void transfer(Account from, Account to, double amount){
      Account first = from;
      Account second = to;
      if (first.compareTo(second) < 0) {
          // Swap them
          first = to;
          second = from;
      }
      synchronized(first){
           synchronized(second){
                from.withdraw(amount);
                to.deposit(amount);
           }
      }
 }
于 2012-11-10T23:06:23.060 回答
10

除了锁定顺序的解决方案之外,您还可以通过在执行任何帐户转移之前同步私有静态最终锁定对象来避免死锁。

 class Account{
 double balance;
 int id;
 private static final Object lock = new Object();
  ....




 public static void transfer(Account from, Account to, double amount){
          synchronized(lock)
          {
                    from.withdraw(amount);
                    to.deposit(amount);
          }
     }

该解决方案的问题是私有静态锁限制系统“按顺序”执行传输。

如果每个帐户都有一个 ReentrantLock,则另一个可能是:

private final Lock lock = new ReentrantLock();




public static void transfer(Account from, Account to, double amount)
{
       while(true)
        {
          if(from.lock.tryLock()){
            try { 
                if (to.lock.tryLock()){
                   try{
                       from.withdraw(amount);
                       to.deposit(amount);
                       break;
                   } 
                   finally {
                       to.lock.unlock();
                   }
                }
           }
           finally {
                from.lock.unlock();
           }

           int n = number.nextInt(1000);
           int TIME = 1000 + n; // 1 second + random delay to prevent livelock
           Thread.sleep(TIME);
        }

 }

这种方法不会发生死锁,因为这些锁永远不会被无限期地持有。如果获取了当前对象的锁,但第二个锁不可用,则释放第一个锁,线程在尝试重新获取锁之前休眠一段指定的时间。

于 2012-11-10T23:32:18.300 回答
9

这是一个经典的问题。我看到两种可能的解决方案:

  1. 对帐户进行排序并在 id 低于另一个帐户的帐户上进行同步。这种方法在并发圣经Java Concurrency in Practice第10章中提到。在本书中作者使用系统哈希码来区分帐户。请参阅java.lang.System#identityHashCode
  2. 您提到了第二种解决方案 - 是的,您可以避免嵌套同步块,并且您的代码不会导致死锁。但在这种情况下,处理可能会出现一些问题,因为如果您从第一个帐户中提取资金,则第二个帐户可能会被锁定很长时间,并且您可能需要将资金放回第一个帐户。这不好,因为嵌套同步和锁定两个帐户是更好且更常用的解决方案。
于 2012-11-11T00:10:19.253 回答
5

您还可以为每个帐户(在帐户类中)创建单独的锁,然后在进行事务之前获取两个锁。看一看:

private boolean acquireLocks(Account anotherAccount) {
        boolean fromAccountLock = false;
        boolean toAccountLock = false;
        try {
            fromAccountLock = getLock().tryLock();
            toAccountLock = anotherAccount.getLock().tryLock();
        } finally {
            if (!(fromAccountLock && toAccountLock)) {
                if (fromAccountLock) {
                    getLock().unlock();
                }
                if (toAccountLock) {
                    anotherAccount.getLock().unlock();
                }
            }
        }
        return fromAccountLock && toAccountLock;
    }

获得两把锁后,您可以进行转移而无需担心安全问题。

    public static void transfer(Acc from, Acc to, double amount) {
        if (from.acquireLocks(to)) {
            try {
                from.withdraw(amount);
                to.deposit(amount);
            } finally {
                from.getLock().unlock();
                to.getLock().unlock();
            }
        } else {
            System.out.println(threadName + " cant get Lock, try again!");
            // sleep here for random amount of time and try do it again
            transfer(from, to, amount);
        }
    }
于 2014-10-07T17:26:55.777 回答
0

这是所述问题的解决方案。

import java.util.Random;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class FixDeadLock1 {

    private class Account {

        private final Lock lock = new ReentrantLock();

        @SuppressWarnings("unused")
        double balance;
        @SuppressWarnings("unused")
        int id;

        public Account(int id, double balance) {
            this.balance = balance;
            this.id = id;
        }

        void withdraw(double amount) {
            this.balance -= amount;
        }

        void deposit(double amount) {
            balance += amount;
        }
    }

    private class Transfer {

        void transfer(Account fromAccount, Account toAccount, double amount) {
            /*
             * synchronized (fromAccount) { synchronized (toAccount) {
             * fromAccount.withdraw(amount); toAccount.deposit(amount); } }
             */

            if (impendingTransaction(fromAccount, toAccount)) {
                try {
                    System.out.format("Transaction Begins from:%d to:%d\n",
                            fromAccount.id, toAccount.id);
                    fromAccount.withdraw(amount);
                    toAccount.deposit(amount);
                } finally {
                    fromAccount.lock.unlock();
                    toAccount.lock.unlock();
                }

            } else {
                 System.out.println("Unable to begin transaction");
            }

        }

        boolean impendingTransaction(Account fromAccount, Account toAccount) {

            Boolean fromAccountLock = false;
            Boolean toAccountLock = false;

            try {
                fromAccountLock = fromAccount.lock.tryLock();
                toAccountLock = toAccount.lock.tryLock();
            } finally {
                if (!(fromAccountLock && toAccountLock)) {
                    if (fromAccountLock) {
                        fromAccount.lock.unlock();
                    }
                    if (toAccountLock) {
                        toAccount.lock.unlock();
                    }
                }
            }

            return fromAccountLock && toAccountLock;
        }

    }

    private class WrapperTransfer implements Runnable {
        private Account fromAccount;
        private Account toAccount;
        private double amount;

        public WrapperTransfer(Account fromAccount,Account toAccount,double amount){
            this.fromAccount = fromAccount;
            this.toAccount = toAccount; 
            this.amount = amount;
        }

        public void run(){
            Random random = new Random();
            try {
                int n = random.nextInt(1000);
                int TIME = 1000 + n; // 1 second + random delay to prevent livelock
                Thread.sleep(TIME);
            } catch (InterruptedException e) {}
            new Transfer().transfer(fromAccount, toAccount, amount);
        }

    }

    public void initiateDeadLockTransfer() {
        Account from = new Account(1, 1000);
        Account to = new Account(2, 300);       
        new Thread(new WrapperTransfer(from,to,200)).start();
        new Thread(new WrapperTransfer(to,from,300)).start();
    }

    public static void main(String[] args) {
        new FixDeadLock1().initiateDeadLockTransfer();
    }

}
于 2018-01-16T17:04:36.527 回答
-1

您必须满足三个要求:

  1. 始终将一个帐户的内容减少指定的数量。
  2. 以指定的数量持续增加另一个帐户的内容。
  3. 如果以上其中一项成功,则另一项也必须成功。

您可以通过使用Atomics来实现 1. 和 2. ,但您将不得不使用其他的东西,double因为没有AtomicDouble. AtomicLong可能是你最好的选择。

所以你剩下第三个要求——如果一个成功,另一个必须成功。有一种简单的技术可以很好地与原子配合使用,那就是使用getAndAdd方法。

class Account {
  AtomicLong balance = new AtomicLong ();
}

...
Long oldDebtor = null;
Long oldCreditor = null;
try {
  // Increase one.
  oldDebtor = debtor.balance.getAndAdd(value);
  // Decrease the other.
  oldCreditor = creditor.balance.gtAndAdd(-value);
} catch (Exception e) {
  // Most likely (but still incredibly unlikely) InterruptedException but theoretically anything.
  // Roll back
  if ( oldDebtor != null ) {
    debtor.getAndAdd(-value);
  }
  if ( oldCreditor != null ) {
    creditor.getAndAdd(value);
  }
  // Re-throw after cleanup.
  throw (e);
}
于 2012-11-11T00:35:32.390 回答