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多线程15-线程案例

时间:2014-06-24 09:17:22      阅读:307      评论:0      收藏:0      [点我收藏+]

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1. 案例一  

       现有的程序代码模拟产生了16个日志对象,并且需要运行16秒才能打印完这些日志,请在程序中增加4个线程去调用parseLog()方法来分别打印这16个日志对象,

程序只需要运行4秒即可打印完这些日志对象,代码如下: 

 

package read;
    
    public class Test {
        
        public static void main(String[] args){
            
            System.out.println("begin:"+(System.currentTimeMillis()/1000));
       
            for(int i=0;i<16;i++){  //这行代码不能改动
                final String log = ""+(i+1);//这行代码不能改动
                {
                         Test.parseLog(log);
                }
            }
        }
        
        //parseLog方法内部的代码不能改动
        public static void parseLog(String log){
            System.out.println(log+":"+(System.currentTimeMillis()/1000));
            
            try {
                Thread.sleep(1000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }        
        }
        
    }

 

解答:

 

package read;

import java.util.concurrent.ArrayBlockingQueue;
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
    
    public class Test {
        
        public static void main(String[] args) throws Exception{
            ExecutorService threadPool = Executors.newFixedThreadPool(4) ;
            final BlockingQueue<String> queue = new  ArrayBlockingQueue<String>(16) ;  //队列长度为16 
            for(int i = 1 ;i<=4 ;i++){
                Runnable runnable = new Runnable() {
                    public void run() {
                        try {
                            for(int i = 0 ;i<4;i++)
                            Test.parseLog(queue.take()) ;
                            
                            if(queue.size()==0){
                                System.out.println("end:"+(System.currentTimeMillis()/1000));
                            }
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                    }
                };
                
                threadPool.execute(runnable) ;
                
            }
            
            
            System.out.println("begin:"+(System.currentTimeMillis()/1000));
            for(int i=0;i<16;i++){  //这行代码不能改动
                final String log = ""+(i+1);//这行代码不能改动
                {
                     //    Test.parseLog(log);
                    queue.put(log) ; 
                }
            }
        }
        
        //parseLog方法内部的代码不能改动
        public static void parseLog(String log){
            System.out.println(log+":"+(System.currentTimeMillis()/1000));
            
            try {
                Thread.sleep(1000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }        
        }
        
    }

 

 

2.案例二

   现成程序中的Test类中的代码在不断地产生数据,然后交给TestDo.doSome()方法去处理,就好像生产者在不断地产生数据,消费者在不断消费数据。请将程序改造成有10个线程来消费生成者产生

数据,这些消费者都调用TestDo.doSome()方法去进行处理,故每个消费者都需要一秒才能处理完,程序应保证这些消费者线程依次有序地消费数据,只有上一个消费者消费完后,下一个消费者才能消

费数据,下一个消费者是谁都可以,但要保证这些消费者线程拿到的数据是有顺序的

   

package queue;
    
    public class Test {
    
        public static void main(String[] args) {
            
            System.out.println("begin:"+(System.currentTimeMillis()/1000));
            for(int i=0;i<10;i++){  //这行不能改动
                String input = i+"";  //这行不能改动
                String output = TestDo.doSome(input);
                System.out.println(Thread.currentThread().getName()+ ":" + output);
            }
        }
    }
    
    //不能改动此TestDo类
    class TestDo {
        public static String doSome(String input){
            
            try {
                Thread.sleep(1000);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            String output = input + ":"+ (System.currentTimeMillis() / 1000);
            return output;
        }
    }

 

解答:

package queue;

import java.util.concurrent.Semaphore;
import java.util.concurrent.SynchronousQueue;

public class Test {

    public static void main(String[] args) {
        final Semaphore semaphore = new Semaphore(1);
        final SynchronousQueue<String> queue = new SynchronousQueue<String>();
        for(int i=0;i<10;i++){
            new Thread(new Runnable(){
                @Override
                public void run() {    
                    try {
                        semaphore.acquire();
                        String input = queue.take();
                        String output = TestDo.doSome(input);
                        System.out.println(Thread.currentThread().getName()+ ":" + output);
                        semaphore.release();
                    } catch (InterruptedException e) {
                        // TODO Auto-generated catch block
                        e.printStackTrace();
                    }    
                }
            }).start();
        }
        
        System.out.println("begin:"+(System.currentTimeMillis()/1000));
        for(int i=0;i<10;i++){  //这行不能改动
            String input = i+"";  //这行不能改动
            try {
                queue.put(input);
            } catch (InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
    }
}

//不能改动此TestDo类
class TestDo {
    public static String doSome(String input){
        
        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        String output = input + ":"+ (System.currentTimeMillis() / 1000);
        return output;
    }
}

 

3.案例三

    

现有程序同时启动了4个线程去调用TestDo.doSome(key, value)方法,由于TestDo.doSome(key, value)方法内的代码是先暂停1秒,然后再输出以秒为单位的当前时间值,所以,会打印出4个相同的时间值,如下所示:
4:4:1258199615
1:1:1258199615
3:3:1258199615
1:2:1258199615
请修改代码,如果有几个线程调用TestDo.doSome(key, value)方法时,传递进去的key相等(equals比较为true),则这几个线程应互斥排队输出结果,即当有两个线程的key都是"1"时,它们中的一个要比另外其他线程晚1秒输出结果,如下所示:
4:4:1258199615
1:1:1258199615
3:3:1258199615
1:2:1258199616
总之,当每个线程中指定的key相等时,这些相等key的线程应每隔一秒依次输出时间值(要用互斥),如果key不同,则并行执行(相互之间不互斥)。原始代码如下:

 

package syn;

    //不能改动此Test类    
    public class Test extends Thread{
        
        private TestDo testDo;
        private String key;
        private String value;
        
        public Test(String key,String key2,String value){
            this.testDo = TestDo.getInstance();
            /*常量"1"和"1"是同一个对象,下面这行代码就是要用"1"+""的方式产生新的对象,
            以实现内容没有改变,仍然相等(都还为"1"),但对象却不再是同一个的效果*/
            this.key = key+key2; 
            this.value = value;
        }


        public static void main(String[] args) throws InterruptedException{
            Test a = new Test("1","","1");
            Test b = new Test("1","","2");
            Test c = new Test("3","","3");
            Test d = new Test("4","","4");
            System.out.println("begin:"+(System.currentTimeMillis()/1000));
            a.start();
            b.start();
            c.start();
            d.start();

        }
        
        public void run(){
            testDo.doSome(key, value);
        }
    }

    class TestDo {

        private TestDo() {}
        private static TestDo _instance = new TestDo();    
        public static TestDo getInstance() {
            return _instance;
        }

        public void doSome(Object key, String value) {
    
            // 以大括号内的是需要局部同步的代码,不能改动!
            {
                try {
                    Thread.sleep(1000);
                    System.out.println(key+":"+value + ":"
                            + (System.currentTimeMillis() / 1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
        }

    }

 

 解答:

 

package syn;

import java.util.ArrayList;
import java.util.Iterator;
import java.util.concurrent.CopyOnWriteArrayList;

//不能改动此Test类    
public class Test extends Thread{
    
    private TestDo testDo;
    private String key;
    private String value;
    
    public Test(String key,String key2,String value){
        this.testDo = TestDo.getInstance();
        /*常量"1"和"1"是同一个对象,下面这行代码就是要用"1"+""的方式产生新的对象,
        以实现内容没有改变,仍然相等(都还为"1"),但对象却不再是同一个的效果*/
        this.key = key+key2; 
/*        a = "1"+"";
        b = "1"+""
*/
        this.value = value;
    }


    public static void main(String[] args) throws InterruptedException{
        Test a = new Test("1","","1");
        Test b = new Test("1","","2");
        Test c = new Test("3","","3");
        Test d = new Test("4","","4");
        System.out.println("begin:"+(System.currentTimeMillis()/1000));
        a.start();
        b.start();
        c.start();
        d.start();

    }
    
    public void run(){
        testDo.doSome(key, value);
    }
}

class TestDo {

    private TestDo() {}
    private static TestDo _instance = new TestDo();    
    public static TestDo getInstance() {
        return _instance;
    }

    //private ArrayList keys = new ArrayList();
    private CopyOnWriteArrayList keys = new CopyOnWriteArrayList();
    public void doSome(Object key, String value) {
        Object o = key;
        if(!keys.contains(o)){
            keys.add(o);
        }else{

            for(Iterator iter=keys.iterator();iter.hasNext();){
                try {
                    Thread.sleep(20);
                } catch (InterruptedException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
                Object oo = iter.next();
                if(oo.equals(o)){
                    o = oo;
                    break;
                }
            }
        }
        synchronized(o)
        // 以大括号内的是需要局部同步的代码,不能改动!
        {
            try {
                Thread.sleep(1000);
                System.out.println(key+":"+value + ":"
                        + (System.currentTimeMillis() / 1000));
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }

}

 

 

 

 

 

 

 

 

    

 

 

 

 

 

 

 

 

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多线程15-线程案例

标签:style   class   blog   code   java   tar   

原文地址:http://www.cnblogs.com/liaokailin/p/3799324.html

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