多线程如何设置2个线程执行顺序

今天药忘吃喽~ 2024-03-17 09:47 40阅读 0赞

### 正常情况下 ###

public class TaskOne extends Thread {
        
        @Override
        public void run() {
        
            System.out.println(Thread.currentThread().getName() + " → " + " Start Time = " + new Date());
            processCommand();
            System.out.println(Thread.currentThread().getName() + " → " + " End   Time = " + new Date());
        }
    
        private void processCommand() {
        
            try {
        
                Thread.sleep(3000);
            } catch (InterruptedException e) {
        
                e.printStackTrace();
            }
        }
    }

public class TaskTwo extends Thread {
        
        @Override
        public void run() {
        
            System.out.println(Thread.currentThread().getName() + " → " + " Start Time = " + new Date());
            processCommand();
            System.out.println(Thread.currentThread().getName() + " → " + " End   Time = " + new Date());
        }
    
        private void processCommand() {
        
            try {
        
                Thread.sleep(3000);
            } catch (InterruptedException e) {
        
                e.printStackTrace();
            }
        }
    }

TaskOne one = new TaskOne();
    TaskTwo two = new TaskTwo();
    one.start();
    two.start();

#### 执行结果 ####

<table> 
 <thead> 
  <tr> 
   <th align="left">第1次执行</th> 
   <th align="left">第2次执行</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td align="left">Thread-0 → Start Time = Mon May 22 21:02:57 CST 2023</td> 
   <td align="left">Thread-1 → Start Time = Mon May 22 21:02:29 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → Start Time = Mon May 22 21:02:57 CST 2023</td> 
   <td align="left">Thread-0 → Start Time = Mon May 22 21:02:29 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-0 → End Time = Mon May 22 21:03:01 CST 2023</td> 
   <td align="left">Thread-0 → End Time = Mon May 22 21:02:32 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → End Time = Mon May 22 21:03:01 CST 2023</td> 
   <td align="left">Thread-1 → End Time = Mon May 22 21:02:32 CST 2023</td> 
  </tr> 
 </tbody> 
</table>

#### 引入问题 ####

**如果这样写的话线程`one`和线程`two`是交替执行的,但是不一定谁先启动。**

### 新设想 ###

**所以`two`线程要在`one`的`run`里面创建，这样保证两个有先后顺序**

/**
     * 保证two在one后面执行
     */
    public class TaskTwoInOne extends Thread {
        
        public void run() {
        
            System.out.println(Thread.currentThread().getName() + " → " + " Start Time = " + new Date());
    
            TaskTwo taskTwo = new TaskTwo();
            taskTwo.start();
            processCommand();
    
            System.out.println(Thread.currentThread().getName() + " → " + " End   Time = " + new Date());
        }
    
        private void processCommand() {
        
            try {
        
                Thread.sleep(3000);
            } catch (InterruptedException e) {
        
                e.printStackTrace();
            }
        }
    }

TaskTwoInOne one = new TaskTwoInOne();
    one.start();

#### 执行结果 ####

<table> 
 <thead> 
  <tr> 
   <th align="left">第1次执行</th> 
   <th align="left">第2次执行</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td align="left">Thread-0 → Start Time = Mon May 22 21:11:56 CST 2023</td> 
   <td align="left">Thread-0 → Start Time = Mon May 22 21:13:17 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → Start Time = Mon May 22 21:11:56 CST 2023</td> 
   <td align="left">Thread-1 → Start Time = Mon May 22 21:13:17 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-0 → End Time = Mon May 22 21:11:59 CST 2023</td> 
   <td align="left">Thread-1 → End Time = Mon May 22 21:13:20 CST 2023</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → End Time = Mon May 22 21:11:59 CST 2023</td> 
   <td align="left">Thread-0 → End Time = Mon May 22 21:13:20 CST 2023</td> 
  </tr> 
 </tbody> 
</table>

#### 引入问题 ####

**以上只是保证，线程`one`比线程`two`先开始执行，后面就开始争抢资源不一定谁先结束**。

**如何保证线程`two`执行完了，再执行线程`one`呢？**

### 引入Join() ###

public class TaskTwoJoinOne extends Thread {
        
        public void run() {
        
            System.out.println(Thread.currentThread().getName() + " → " + " Start Time = " + new Date());
    
            TaskTwo two = new TaskTwo();
            two.start();
            try {
        
                two.join(); // 线程two加入到线程one里,two执行完one才开始执行
            } catch (InterruptedException e) {
        
                e.printStackTrace();
            }
            System.out.println(Thread.currentThread().getName() + " → " + " End   Time = " + new Date());
        }
    }

#### 执行结果 ####

<table> 
 <thead> 
  <tr> 
   <th align="left">第1次执行</th> 
   <th align="left">结果分析</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td align="left"><strong>前置条件</strong>：<code>two.join();</code></td> 
   <td align="left">线程<code>two</code>加入到线程<code>one</code>里，<code>two</code> 执行完 <code>one</code> 才开始执行</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-0 → Start Time = Mon May 22 21:24:24 CST 2023</td> 
   <td align="left">线程<code>one</code> 开始</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → Start Time = Mon May 22 21:24:24 CST 2023</td> 
   <td align="left">线程<code>two</code> <code>join</code></td> 
  </tr> 
  <tr> 
   <td align="left">Thread-1 → End Time = Mon May 22 21:24:27 CST 2023</td> 
   <td align="left">线程<code>two</code> 执行完毕</td> 
  </tr> 
  <tr> 
   <td align="left">Thread-0 → End Time = Mon May 22 21:24:27 CST 2023</td> 
   <td align="left">线程<code>one</code> 执行完毕</td> 
  </tr> 
  <tr> 
   <td align="left"></td> 
   <td align="left"></td> 
  </tr> 
  <tr> 
   <td align="left"><strong>前置条件</strong>：<code>two.join(2000);</code></td> 
   <td align="left">线程<code>two</code>加入到线程<code>one</code>里,就等线程<code>two</code> 2000ms,超过就不等了</td> 
  </tr> 
  <tr> 
   <td align="left"></td> 
   <td align="left"></td> 
  </tr> 
  <tr> 
   <td align="left"><strong>前置条件</strong>：<code>two.join(6000);</code></td> 
   <td align="left">最长需等6000ms,如果线程<code>two</code>执行完不需要6000ms，则直接执行线程<code>one</code>内容</td> 
  </tr> 
 </tbody> 
</table>

#### 引入问题 ####

**这样就太给线程`two`特权了，线程`two`不执行完，线程`one`就一直等着！！！**

--------------------

### 引入 join(1000) ###

public class TaskTwoJoinOne extends Thread {
        
        public void run() {
        
            System.out.println(Thread.currentThread().getName() + " → " + " Start Time = " + new Date());
    
            TaskTwo two = new TaskTwo();
            two.start();
            try {
        
                two.join(1000);// 线程two加入到线程one里,就等线程two2000毫秒,超过就不等了
    
                // two.join(6000);// 最长需等6000ms,如果线程two执行完不需要6000ms，则可以继续执行线程one的内容
            } catch (InterruptedException e) {
        
                e.printStackTrace();
            }
            System.out.println(Thread.currentThread().getName() + " → " + " End   Time = " + new Date());
        }
    }