[英]How do I ensure that one thread always run ahead of other in java? Given two threads. I need to make sure that Thread 1 always leads Thread 2
给定两个线程。 线程 1:从 1 打印到 100 线程 2:从 1 打印到 100 我需要确保线程 1 从不打印线程 2 尚未打印的任何数字。
请帮我解决一下这个。 提前致谢。 (——在我对维森科技的采访中被问到)
我尝试了以下方法。
package threadTutor;
public class ThreadLeader {
public static void main(String[] args) throws InterruptedException {
Leader lead = new Leader(100);
Follower foll = new Follower(100);
foll.start();
lead.start();
foll.join();
lead.join();
}
}
class Leader extends Thread{
static int start;
static int end;
static int pos;
Leader(int end){
Leader.end = end;
}
public void run() {
for(int i=start;i<end;i++) {
System.out.println("Leader is printing "+i);
Leader.pos=i;
}
}
}
class Follower extends Thread{
static int start;
static int end;
static int pos;
Follower(int end){
Follower.end = end;
}
public void run() {
while(true) {
if(Follower.pos<=Leader.pos) {
System.out.println("Follower is printing "+pos);
pos++;
}
if(Follower.pos==Follower.end) break;
}
}
}
但是当然,无限的 while 循环不是做事的好方法。 请提供更好的方法。
这是一个消费者-生产者问题。 以下是阻塞队列的实现。 其他选项: https://www.geeksforgeeks.org/producer-consumer-solution-using-threads-java/
import java.util.concurrent.BlockingQueue;
import java.util.concurrent.LinkedBlockingQueue;
public class ThreadLeader {
public static void main(String[] args) throws InterruptedException {
BlockingQueue<Integer> queue = new LinkedBlockingQueue<>();
Leader lead = new Leader(queue, 100);
Follower foll = new Follower(queue, 100);
foll.start();
lead.start();
foll.join();
lead.join();
}
}
class Leader extends Thread {
private BlockingQueue<Integer> queue;
private int num;
Leader(BlockingQueue<Integer> queue, int num) {
this.queue = queue;
this.num = num;
}
public void run() {
for (int i = 1; i <= num; i++) {
System.out.println("Leader is printing " + i);
queue.offer(i);
}
}
}
class Follower extends Thread {
private BlockingQueue<Integer> queue;
private int num;
Follower(BlockingQueue<Integer> queue, int num) {
this.queue = queue;
this.num = num;
}
public void run() {
for (int i = 0; i < num; i++){
try {
System.out.println("Follower is printing " + queue.take());
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
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