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。
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;
}
}
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();
}
}
}
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。
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;
}
}
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();
}
}
}
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
![](https://img2018.cnblogs.com/blog/1475083/201812/1475083-20181223131656952-439937296.png)
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。