生產者消費者問題——Java方案

最近有個同學讓我幫忙寫下多線程，由於不經常使用Java，對多線程忘了差不多了，於是就想來複習一下。然後就自然想到了著名的生產者-消費者問題，於是就寫了個小程序來練下手o(∩_∩)o

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package stwolf.hustbaidu.java.learn; public class ProducerConsumer { public static final int N = 10; static Producer myProducer = new Producer(); static Consumer myConsumer = new Consumer(); static Monitor myMonitor = new Monitor(); public static void main(String args[]) { myProducer.start(); myConsumer.start(); } static class Producer extends Thread //生產者類 { @Override public void run() { int item; while (true) { item = produce_item(); myMonitor.insert(item); } } private int produce_item() //實際生產活動 { return (int)Math.random(); } } static class Consumer extends Thread //消費者類 { @Override public void run() { int item; while (true) { item = myMonitor.remove(); consume_item(item); } } private void consume_item(int item) //實際生產活動 { } } static class Monitor //管程 { private int[] buffer = new int[N]; private int count = 0, low = 0, high = 0; public synchronized void insert(int val) { if (count == N) //緩衝區滿，休眠 { sleep(); } System.out.println("Produce" + high); buffer[high] = val; high = (high + 1) % N; //設置下一項索引 count++; if (count == 1) //若消費者在休眠，喚醒消費者 { notify(); } } public synchronized int remove() { int val; if (count == 0) ////緩衝區空，休眠 { sleep(); } System.out.println("Consume" + low); val = buffer[low]; low = (low + 1) % N; //設置下一項索引 count--; if (count == N - 1) //若生產者在休眠，喚醒生產者 { notify(); } return val; } private void sleep() { try { wait(); } catch (InterruptedException e) { e.printStackTrace(); } } } }

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程序的部分輸出如下：

Produce0 Consume0 Produce1 Consume1 Produce2 Consume2 Produce3 Consume3 Produce4 Consume4 Produce5 Consume5 Produce6 Consume6 Produce7 Consume7 Produce8 Consume8 Produce9 Consume9 Produce0 Consume0 Produce1 Produce2 Produce3 Produce4 Produce5 Produce6 Produce7 Produce8 Produce9 Produce0 Consume1 Consume2 Consume3 Consume4 Consume5