我正在开发一个带有两个线程的循环缓冲区:消费者和生产者。我正在使用主动等待Thread.yield
。我知道可以使用信号量来做到这一点,但我想要没有信号量的缓冲区。
两者都有一个共享变量:bufferCircular
.
当缓冲区没有充满有用信息时,producer
在数组位置写入数据p
,当有一些有用信息consumer
时,在数组位置读取数据c
。变量nElem
fromBufferCircular
是尚未读取的值数据的数量。
该程序运行良好 9/10 次。然后,有时,它会在屏幕上显示最后一个元素(循环的编号 500)之前陷入无限循环,或者只是不显示任何元素。
我想可能是一个liveLock,但我找不到错误。
共享变量:
public class BufferCircular {
volatile int[] array;
volatile int p;
volatile int c;
volatile int nElem;
public BufferCircular(int[] array) {
this.array = array;
this.p = 0;
this.c = 0;
this.nElem = 0;
}
public void writeData (int data) {
this.array[p] = data;
this.p = (p + 1) % array.length;
this.nElem++;
}
public int readData() {
int data = array[c];
this.c = (c + 1) % array.length;
this.nElem--;
return data;
}
}
生产者线程:
public class Producer extends Thread {
BufferCircular buffer;
int bufferTam;
int contData;
public Productor(BufferCircular buff) {
this.buffer = buff;
this.bufferTam = buffer.array.length;
this.contData = 0;
}
public void produceData() {
this.contData++;
this.buffer.writeData(contData);
}
public void run() {
for (int i = 0; i < 500; i++) {
while (this.buffer.nElem == this.bufferTam) {
Thread.yield();
}
this.produceData();
}
}
}
消费者线程:
public class Consumer extends Thread {
BufferCircular buffer;
int cont;
public Consumer(BufferCircular buff) {
this.buffer = buff;
this.cont = 0;
}
public void consumeData() {
int data = buffer.readData();
cont++;
System.out.println("data " + cont + ": " + data);
}
public void run() {
for (int i = 0; i < 500; i++) {
while (this.buffer.nElem == 0) {
Thread.yield();
}
this.consumeData();
}
}
}
主要:
public class Main {
public static void main(String[] args) {
Random ran = new Random();
int tamArray = ran.nextInt(21) + 1;
int[] array = new int[tamArray];
BufferCircular buffer = new BufferCircular(array);
Producer producer = new Producer (buffer);
Consumer consumer = new Consumer (buffer);
producer.start();
consumer.start();
try {
producer.join();
consumer.join();
} catch (InterruptedException e) {
System.err.println("Error with Threads");
e.printStackTrace();
}
}
}
欢迎任何帮助。