       這是以前為一師妹做的一機試題，雖后來因為她編程底子太弱，沒能獲得那份工作，拿在這里與大家共享之。
       題目如下：
      在生產(chǎn)者／消費者模型中，生產(chǎn)者Producer負責生產(chǎn)數(shù)據(jù)，而消費者Consumer負責使用數(shù)據(jù)。多個生產(chǎn)者線程會在同一時間運行，生產(chǎn)數(shù)據(jù)，并放到內(nèi)存中一個共享的區(qū)域。期間，多個消費者線程讀取內(nèi)存共享區(qū)，消費里面的數(shù)據(jù)。
要求：
     1. 針對上面的場景，請創(chuàng)建2個類，一個叫Producer,一個叫Consumer.
     2. Producer類繼承Thread類，并實現(xiàn)把數(shù)據(jù)放到內(nèi)存共享區(qū)的功能，這個功能要求是線程安全的。在個Producer類中的run方法中，循環(huán)20次，每次把一個整數(shù)放到內(nèi)存共享區(qū)中。
     3. Consumer類也繼承Thread類，并實現(xiàn)在沒有沖突的情況下，從內(nèi)存共享區(qū)中獲取數(shù)據(jù)，并在標準輸出設備中打印輸出。輸出的格式為：Consumer thread X retrieved integer Y.
     4. 最后，創(chuàng)建一個main class,創(chuàng)建10個Procuder線程和4個消費者線程并啟動這些線程。
     5. 要求有效代碼行數(shù)盡量少，最好不超過100行。
     我大概的做了一下，以下是我的代碼實現(xiàn)：
     一. 消費者類

/**

* 消費者類

* @author Amigo Xie(xiexingxing1121@126.com)

class Consumer extends Thread {

private Conn conn;

private int consumerNumber;

public Consumer(Conn conn1, int conNum) {

conn = conn1;

consumerNumber = conNum;

}

public void run() {

for (int i = 0; i < 50; i++) {

System.out.println("Consumer thread " + consumerNumber + " retrieved integer: " + conn.read()); //

try {

sleep((int) (Math.random() * 2000));

} catch (InterruptedException e) {

e.printStackTrace();

}

      二. 生產(chǎn)者類

/**

* 生產(chǎn)者類

* @author Amigo Xie(xiexingxing1121@126.com)

class Producer extends Thread {

private Conn conn;

public Producer(Conn conn1) {

conn = conn1;

}

public void run() {

for (int i = 0; i < 20; i++) {

conn.add(i);

}

三. 線程通信類

/**

* 線程通信類

* @author Amigo Xie(xiexingxing1121@126.com)

class Conn {

private int buffer[] = new int[200]; //10個Procuder線程，需存放200個變量

private int next = 0; //Flags to keep track of our int buffer status

private boolean isFull = false;

private boolean isEmpty = true;

/**

* method to read int

* @return

public synchronized int read() {

while (isEmpty == true) { //We can't read if there is nothing in our int buffer

try {

wait();//we'll exit this when isEmpty turns false

}catch (InterruptedException e) {

e.printStackTrace();

}

next--; //decrement the count,since we're going to read one int

if (next == 0) {

isEmpty = true; //Did we read the last letter?

}

isFull = false;

notify();

return (buffer[next]);//return the int to the thread that is reading

}

/**

* method to add integer to the buffer

* @param number

public synchronized void add(int number) {

while (isFull == true ) { //Wait around until there's room to add another letter

try {

wait();//This will exit when isFull turns false

}catch (InterruptedException e) {

e.printStackTrace();

}

next++; //add the integer to the next available spot buffer[next]=number;Change the next available spot

if (next == 200) {

isFull = true; //Are we full?

} else {

buffer[next] = number;

}

isEmpty =false;

notify();

}

四. 測試類

/**

* 測試類

* @author Amigo Xie(xiexingxing1121@126.com)

public class ProducerAndConsumerTest {

/**

* @param args

public static void main(String[] args) {

Conn conn = new Conn();

Producer pro1 = new Producer(conn);

Producer pro2 = new Producer(conn);

Producer pro3 = new Producer(conn);

Producer pro4 = new Producer(conn);

Producer pro5 = new Producer(conn);

Producer pro6 = new Producer(conn);

Producer pro7 = new Producer(conn);

Producer pro8 = new Producer(conn);

Producer pro9 = new Producer(conn);

Producer pro10 = new Producer(conn);

Consumer consumer1 = new Consumer(conn, 1);

Consumer consumer2 = new Consumer(conn, 2);

Consumer consumer3 = new Consumer(conn, 3);

Consumer consumer4 = new Consumer(conn, 4);

pro1.start();

pro2.start();

pro3.start();

pro4.start();

pro5.start();

pro6.start();

pro7.start();

pro8.start();

pro9.start();

pro10.start();

consumer1.start();

consumer2.start();

consumer3.start();

consumer4.start();

}

近來發(fā)現(xiàn)，去看以前的代碼，都覺得有點傻傻滴，現(xiàn)在的話如果去實現(xiàn)大抵會有所改進，我在進步中嗎？啊哈。

posted on 2007-04-11 21:52 阿蜜果閱讀(6294) 評論(8) 編輯收藏所屬分類: Java

FeedBack:

# re: 生產(chǎn)者／消費者模型模擬實現(xiàn)[未登錄]

2007-04-11 23:01 | Tony

樓主的測試類可不怎么幽雅,呵呵回復更多評論