小贝子编程

避免僵局例子



我想知道在以下示例中避免死锁的另一种方法是什么。以下示例是一个典型的银行帐户转移僵局问题。在实践中解决哪些更好的方法？

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，如果我将嵌套锁定在转移方法中，是否会解决僵局问题 

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

对帐户进行排序。死锁来自帐户的订购（a，b vs b，a）。

所以尝试：

 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);
           }
      }
 }

除了订购锁的解决方案外，您还可以通过在执行任何帐户传输之前在私人静态最终锁定对象上同步来避免僵局。

 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，私有静态锁将系统限制为执行"顺序"传输。

另一个可以是，如果每个帐户都有一个重新输入：

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);
        }
 }

在这种方法中不会发生僵局，因为这些锁永远不会被无限期地固定。如果获取了当前对象的锁，但第二个锁定不可用，则第一个锁定释放，线程在尝试重新锁定之前的指定时间内入睡。

这是一个经典的问题。我看到了两个可能的解决方案：

  1. 对帐户进行排序并同步，该帐户的ID低于另一个ID。在第10章的实践中，圣经中提到的此方法Java并发。在本书中，作者使用系统哈希代码来区分帐户。请参阅Java.lang.System＃IdentityHashCode。
  2. 您提到了第二个解决方案 - 是的，您可以避免嵌套同步块，并且您的代码不会导致死锁。但是，在这种情况下，处理可能会遇到一些问题，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在任何大量时间内锁定，并且可能需要将资金放回第一个帐户。那不是很好，因为嵌套的同步和两个帐户的锁定是更好，更常用的解决方案。

您还可以为每个帐户（在帐户类中）创建单独的锁，然后在进行交易之前获得两个锁。看看：

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;
    }

获得两个锁后，您可以转移而不必担心安全。

    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);
        }
    }

这是解决问题的解决方案。

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:%dn",
                            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();
    }
}

您必须满足三个要求：

  1. 始终将一个帐户的内容减少到指定金额。
  2. 始终将另一个帐户的内容增加指定金额。
  3. 如果以上一个成功，另一个也必须成功。

您可以实现1.和2.使用原子，但是您必须使用其他double的东西，因为没有AtomicDouble。AtomicLong可能是您最好的选择。

因此，您需要第三个要求 - 如果一个成功，另一个必须成功。有一种简单的技术，可以与原子学出色，并且使用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);
}

相关内容

最新更新


热门标签：

javascript python java c# php android html jquery c++ css ios sql mysql arrays asp.net json python-3.x ruby-on-rails .net sql-server django objective-c excel regex ruby linux ajax iphone xml vba spring asp.net-mvc database wordpress string postgresql wpf windows xcode bash git oracle list vb.net multithreading eclipse algorithm macos powershell visual-studio image forms numpy scala function api selenium