穷吉201771010119

一,理论知识

1.ava通过多线程的并发运行提高系统资源利用率,改善系统性能。

2.假设有两个或两个以上的线程共享 某个对象,每个线程都调用了改变该对象类状态的方法,就会引起的不确定性。

3.多线程并发执行中的问题

◆多个线程相对执行的顺序是不确定的。

◆线程执行顺序的不确定性会产生执行结果的不确定性。

◆在多线程对共享数据操作时常常会产生这种不确定性。

4.多线程并发运行不确定性问题解决方案:引入线程同步机制

二、实验部分

1、实验目的与要求

(1) 掌握线程同步的概念及实现技术; 

(2) 线程综合编程练习

2、实验内容和步骤

实验1:测试程序并进行代码注释。

测试程序1:

l 在Elipse环境下调试教材651页程序14-7,结合程序运行结果理解程序;

l 掌握利用锁对象和条件对象实现的多线程同步技术。

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package 导入;

import java.util.*;
import java.util.concurrent.locks.*;

/**
 * A bank with a number of bank accounts that uses locks for serializing access.
 * @version 1.30 2004-08-01
 * @author Cay Horstmann
 */
public class Bank
{
   private final double[] accounts;//银行运转的基本数据
   private Lock bankLock;//锁对象
   private Condition sufficientFunds;//

   /**
    * Constructs the bank.
    * @param n the number of accounts
    * @param initialBalance the initial balance for each account
    */
   public Bank(int n, double initialBalance)
   {
      accounts = new double[n];
      Arrays.fill(accounts, initialBalance);
      bankLock = new ReentrantLock();
      sufficientFunds = bankLock.newCondition();
   }

   /**
    * Transfers money from one account to another.
    * @param from the account to transfer from
    * @param to the account to transfer to
    * @param amount the amount to transfer
    */
   public void transfer(int from, int to, double amount) throws InterruptedException
   {
      bankLock.lock();
      try
      {//锁对象的引用条件对象
         while (accounts[from] < amount)
           sufficientFunds.await();
         System.out.print(Thread.currentThread());//打印出线程号
         accounts[from] -= amount;
         System.out.printf(" %10.2f from %d to %d", amount, from, to);
         accounts[to] += amount;
         System.out.printf(" Total Balance: %10.2f%n", getTotalBalance());
         sufficientFunds.signal();
      }
      finally
      {
         bankLock.unlock();
      }
   }

   /**
    * Gets the sum of all account balances.
    * @return the total balance
    */
   public double getTotalBalance()
   {
      bankLock.lock();//加锁
      try
      {
         double sum = 0;

         for (double a : accounts)
            sum += a;

         return sum;
      }
      finally
      {
         bankLock.unlock();//解锁
      }
   }

   /**
    * Gets the number of accounts in the bank.
    * @return the number of accounts
    */
   public int size()
   {
      return accounts.length;
   }
}
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package 导入;

/**
 * This program shows how multiple threads can safely access a data structure.
 * @version 1.31 2015-06-21
 * @author Cay Horstmann
 */
public class SynchBankTest
{
   public static final int NACCOUNTS = 100;
   public static final double INITIAL_BALANCE = 1000;
   public static final double MAX_AMOUNT = 1000;
   public static final int DELAY = 10;
   
   public static void main(String[] args)
   {
      Bank bank = new Bank(NACCOUNTS, INITIAL_BALANCE);
      for (int i = 0; i < NACCOUNTS; i++)
      {
         int fromAccount = i;
         Runnable r = () -> {
            try
            {
               while (true)
               {
                  int toAccount = (int) (bank.size() * Math.random());
                  double amount = MAX_AMOUNT * Math.random();
                  bank.transfer(fromAccount, toAccount, amount);
                  Thread.sleep((int) (DELAY * Math.random()));
               }
            }
            catch (InterruptedException e)
            {
            }            
         };
         Thread t = new Thread(r);
         t.start();
      }
   }
}
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测试程序2:

l  在Elipse环境下调试教材655页程序14-8,结合程序运行结果理解程序;

l  掌握synchronized在多线程同步中的应用。

复制代码
package 导入;

import java.util.*;

/**
 * 使用同步原语的具有多个银行帐户的银行
 * @version 1.30 2004-08-01
 * @author Cay Horstmann
 */
public class Bank
{
   private final double[] accounts;

   /**
    * 构建了银行。
    * @param  账户数量
    * @param 每个账户的初始余额
    */
   public Bank(int n, double initialBalance)
   {
      accounts = new double[n];
      Arrays.fill(accounts, initialBalance);
   }

   /**
    * 把钱从一个账户转到另一个账户。
    * @param 从账户转出
    * @param 到账转到
    * @param 转帐金额
    */
   public synchronized void transfer(int from, int to, double amount) throws InterruptedException
   {
      while (accounts[from] < amount)
         wait();//Object类
      System.out.print(Thread.currentThread());
      accounts[from] -= amount;
      System.out.printf(" %10.2f from %d to %d", amount, from, to);
      accounts[to] += amount;
      System.out.printf(" Total Balance: %10.2f%n", getTotalBalance());
      notifyAll();
   }
   /**
    *获取所有帐户余额的总和。
    * @return 总平衡
    */
   public synchronized double getTotalBalance()
   {
      double sum = 0;

      for (double a : accounts)
         sum += a;

      return sum;
   }

   /**
    * 获取银行中的帐户编号。
    * @return 账户数量
    */
   public int size()
   {
      return accounts.length;
   }
}
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package 导入;

/**
 * 这个程序展示了多线程如何安全地访问一个数据结构,使用同步方法。
 * @version 1.31 2015-06-21
 * @author Cay Horstmann
 */
public class SynchBankTest2
{
   public static final int NACCOUNTS = 100;
   public static final double INITIAL_BALANCE = 1000;
   public static final double MAX_AMOUNT = 1000;
   public static final int DELAY = 10;

   public static void main(String[] args)
   {
      Bank bank = new Bank(NACCOUNTS, INITIAL_BALANCE);
      for (int i = 0; i < NACCOUNTS; i++)
      {
         int fromAccount = i;
         Runnable r = () -> {
            try
            {
               while (true)
               {
                  int toAccount = (int) (bank.size() * Math.random());
                  double amount = MAX_AMOUNT * Math.random();
                  bank.transfer(fromAccount, toAccount, amount);
                  Thread.sleep((int) (DELAY * Math.random()));
               }
            }
            catch (InterruptedException e)
            {
            }
         };
         Thread t = new Thread(r);
         t.start();
      }
   }
}
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测试程序3:

l  在Elipse环境下运行以下程序,结合程序运行结果分析程序存在问题;

l  尝试解决程序中存在问题。

class Cbank

{

     private static int s=2000;

     public   static void sub(int m)

     {

           int temp=s;

           temp=temp-m;

          try {

                     Thread.sleep((int)(1000*Math.random()));

                   }

           catch (InterruptedException e)  {              }

                 s=temp;

                 System.out.println("s="+s);

             }

}

 

 

class Customer extends Thread

{

  public void run()

  {

   for( int i=1; i<=4; i++)

     Cbank.sub(100);

    }

 }

public class Thread3

{

 public static void main(String args[])

  {

   Customer customer1 = new Customer();

   Customer customer2 = new Customer();

   customer1.start();

   customer2.start();

  }

}

源程序运行结果

修改后的代码

class Cbank

{

     private static int s=2000;

     public synchronized static void sub(int m)

     {

           int temp=s;

           temp=temp-m;

          try {

      Thread.sleep((int)(1000*Math.random()));

    }

           catch (InterruptedException e)  {              }

           s=temp;

           System.out.println("s="+s);

   }

}

 

 

class Customer extends Thread

{

  public void run()

  {

   for( int i=1; i<=4; i++)

     Cbank.sub(100);

    }

 }

public class Thread3

{

 public static void main(String args[])

  {

   Customer customer1 = new Customer();

   Customer customer2 = new Customer();

   customer1.start();

   customer2.start();

  }

}
结果图:

实验2 编程练习

利用多线程及同步方法,编写一个程序模拟火车票售票系统,共3个窗口,卖10张票,程序输出结果类似(程序输出不唯一,可以是其他类似结果)。

Thread-0窗口售:第1张票

Thread-0窗口售:第2张票

Thread-1窗口售:第3张票

Thread-2窗口售:第4张票

Thread-2窗口售:第5张票

Thread-1窗口售:第6张票

Thread-0窗口售:第7张票

Thread-2窗口售:第8张票

Thread-1窗口售:第9张票

Thread-0窗口售:第10张票

public class Demo {
    public static void main(String[] args) {
        Mythread mythread = new Mythread();
        Thread ticket1 = new Thread(mythread);
        Thread ticket2 = new Thread(mythread);
        Thread ticket3 = new Thread(mythread);
        ticket1.start();
        ticket2.start();
        ticket3.start();
    }
}

class Mythread implements Runnable {
    int ticket = 1;
    boolean flag = true;

    @Override
    public void run() {
        while (flag) {
            try {
                Thread.sleep(500);
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

            synchronized (this) {
                if (ticket <= 10) {
                    System.out.println(Thread.currentThread().getName() + "窗口售:第" + ticket + "张票");
                    ticket++;
                }
                if (ticket > 10) {
                    flag = false;
                }
            }
        }
    }

}
结果图:
实验总结:这次的试验编程题较比以前写的简单的多,对于我来说这门学课完全是一个新的知识以前所没有接触过得。虽然学起来有的代码写不出,有的理解不了。但还是在尽力去理解。尽管不知道学到了多好但至少接触了一门以前一直没接触的知识。
 
原文地址:https://www.cnblogs.com/qiongji/p/10164040.html