- 时间：2021-07-17 22:49 编辑： 来源： 阅读：
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摘要：java解决单缓冲生产者消费者问题示例
经典的生产者消费者问题模拟。此程序模拟最简单情形——单缓冲。为模拟实际情况，consume item和produce item时加了延时，可以通过修改延时模拟不同的生成消费速率。
[code]
[/co/**
* single buffer consumer-producer problem.
* by xu(xusiwei1236@163.com).
* */
public class ConsumerProducer {
static Object buffer = null;
static Object mutex = new Object();
static Object condConsumer = new Object();
static Object condProducer = new Object();
public static void main(String[] args) {
Thread producer = new Thread() {
public void run() {
// for(int i=0; i<10; i++) {
for(int i=0; ; i++) {
// produce item.
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
}
String item = new String("item-" + i);
System.out.println("[producer] produced " + item);
// wait for buffer empty.
synchronized (condProducer) {
while(buffer != null) {
try {
condProducer.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
// put item to buffer.
synchronized (mutex) {
buffer = item;
System.out.println("[producer] put " + item + " to buffer.");
}
// notify consumers.
synchronized (condConsumer) {
condConsumer.notify();
}
}
}
};
Thread consumer = new Thread() {
public void run() {
// for(int i=0; i<10; i++) {
for( ; ; ) {
// wait for item come.
synchronized (condConsumer) {
while( buffer == null ) {
try {
condConsumer.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
// get item from buffer.
String item = null;
synchronized (mutex) {
item = (String)buffer;
buffer = null;
System.out.println(" [consumer] get " + item + " from buffer.");
}
// consume item.
try {
Thread.sleep(500);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println(" [consumer] comsumed " + item);
// notify producers.
synchronized (condProducer) {
condProducer.notify();
}
}
}
};
consumer.start();
producer.start();
}
}de]
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