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201871010113-刘兴瑞《面向对象程序设计(java)》第十七周学习总结

作者:互联网

 

 

项目

内容

这个作业属于哪个课程

<任课教师博客主页链接>https://www.cnblogs.com/nwnu-daizh/

这个作业的要求在哪里

<作业链接地址>https://www.cnblogs.com/nwnu-daizh/p/12073034.html

作业学习目标

(1) 理解和掌握线程的优先级属性及调度方法;

(2) 掌握线程同步的概念及实现技术;

(3) Java线程综合编程练习

 

第一部分:总结线程同步技术

 

1.Java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。

5.(1)锁对象与条件对象

           用ReentrantLock保护代码块的基本结构如下:

           myLock.lock();

           try { critical section }

           finally{

           myLock.unlock();

   }

   (2)synchronized关键字

          synchronized关键字作用:

         ➢ 某个类内方法用synchronized 修饰后,该方 法被称为同步方法;

         ➢ 只要某个线程正在访问同步方法,其他线程 欲要访问同步方法就被阻塞,直至线程从同 步方法返回前唤醒被阻塞线程,其他线程方 可能进入同步方法。

第二部分: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。

实验代码注释如下:

Bank.java:

package synch;

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;
   private Lock bankLock;
   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();//在等待条件前,锁必须由当前线程保持。
   }

   /**
    * 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();//获取锁
      try
      {
         while (accounts[from] < amount)
            sufficientFunds.await();//造成当前线程在接到信号或被中断之前一直处于等待状态。
         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());
         sufficientFunds.signalAll();//如果所有的线程都在等待此条件,则唤醒所有线程
      }
      finally
      {
         bankLock.unlock();//释放锁。 
      }
   }

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

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

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

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

  SynchBankTest.java:

package synch;

/**
 * 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();
      }
   }
}

运行结果如下:

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。

实验代码注释如下:

Bank.java:

 

package synch2;

import java.util.*;

/**
 * A bank with a number of bank accounts that uses synchronization primitives.
 * @version 1.30 2004-08-01
 * @author Cay Horstmann
 */
public class Bank
{
   private final double[] accounts;

   /**
    * 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);
   }

   /**
    * 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 synchronized void transfer(int from, int to, double amount) throws InterruptedException
   {
      while (accounts[from] < amount)
         wait();//在其他线程调用此对象的 notify() 方法或 notifyAll() 方法前,导致当前线程等待
      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();//唤醒在此对象监视器上等待的所有线程
   }

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

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

      return sum;
   }

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

 

SynchBankTest2.java:

package synch2;

/**
 * This program shows how multiple threads can safely access a data structure,
 * using synchronized methods.
 * @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();
      }
   }
}

代码运行结果如下:

 

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();

  }

}

代码运行结果如图:

运行结果显示两个线程各自运行各自的:并没有实现需要的两个线程一起每次减一百,减八次。

修改代码如下:

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();

  }

}

运行结果如下:

实验2:结对编程练习包含以下4部分(10分)

1)   程序设计思路简述;

2)   符合编程规范的程序代码;

3)   程序运行功能界面截图;

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

 代码如下:

import javax.swing.plaf.SliderUI;

public class shou {
public static void main(String[] args) {
 Mythread mythread=new Mythread();
 Thread t1=new Thread(mythread);
 Thread t2=new Thread(mythread);
 Thread t3=new Thread(mythread);
 t1.start();
 t2.start();
 t3.start();
}
}
 /*
    new Thread() {
        @Override
        public void run() {
            System.out.println();
    };
}.start();
}
}*/

class Mythread implements Runnable{
int t=1;
boolean flag=true;
    @Override
    public void run() {
        while(flag) {
            try {
        
            Thread.sleep(500);
            }
            catch (InterruptedException e) {
                // TODO: handle exception
                e.printStackTrace();
            }
            synchronized (this) {
            if(t<=10) {
                System.out.println(Thread.currentThread().getName()+"窗口售:第"+t+"张票");
                t++;
            }
            if(t<0) {
                flag=false;
            }
        }
    }
    }
}

    

运行结果如图:

实验总结;通过本次实验,我学习到了线程同步的概念,以及如何处理。通过本学期的学习,由刚开始的新手小白,对Java一无所知,到通过大量的练习慢慢对Java编程有所熟悉,虽然现在还不是很熟练,但在课程结束后仍需继续关注学习Java的知识及编程。

标签:java,201871010113,Thread,int,刘兴瑞,线程,double,accounts,public
来源: https://www.cnblogs.com/lxr0/p/12074639.html