run调用其他函数中不会切换到其他线程

问题描述：

如果Thread.sleep()写在Weather类中，会出现输出生成温度后等待较长时间，之后快速输出读取温度和生成温度。这是不是说明，run函数调用其他函数过程中不会切换到其他线程

相关代码：


package thread;

import java.util.Random;

public class Weather {
    public int temperature;
    public int humidity;
    public boolean flag = false;

    public synchronized void set() {
        if(flag){
            try {

                wait();
            } catch (InterruptedException interruptedException) {
                interruptedException.printStackTrace();
            }
        }
        Random r = new Random();
        temperature = r.nextInt(41);
        humidity = r.nextInt(41);
        System.out.println("生成温度:" + temperature + ",湿度:" + humidity);
        flag = true;
        notifyAll();

    }

    public synchronized void get() {
        if(!flag){
            try {

                wait();
            } catch (InterruptedException interruptedException) {
                interruptedException.printStackTrace();
            }
        }
        System.out.println("读取温度:" + temperature + ",湿度:" + humidity);
        flag = false;
        notifyAll();

    }
}

package thread;

public class ReadWeather implements Runnable{
    public Weather e;
    ReadWeather(Weather e){
        this.e = e;
    }

    @Override
    public void run() {
        int i = 0;
        while(i++<100){

            e.get();
            try {
                Thread.sleep(100);
            } catch (InterruptedException interruptedException) {
                interruptedException.printStackTrace();
            }
        }

    }
}

package thread;

public class GenerateWeather implements Runnable{
    public Weather e;
    GenerateWeather(Weather e){
        this.e = e;
    }

    @Override
    public void run() {
        int i = 0;
        while(i++<100){

            e.set();
            try {
                Thread.sleep(1000);
            } catch (InterruptedException interruptedException) {
                interruptedException.printStackTrace();
            }
        }

    }
}

慕运维9261880 2021-04-03

源自：Java多线程 5-3 自由编程

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1回答

好帮手慕小尤回答被采纳获得+3积分 2021-04-03 14:14:04

同学你好，并不是run调用就保证部切换线程了，而是在代码中调用wait()方法使线程处于等待状态，当调用notifyAll()唤醒线程时，才可以继续执行。并在代码中使用了synchronized关键字，synchronized可以保证同一时刻只有一个线程可以执行某个方法（对象）。从而使线程在执行时，不来回切换。

祝学习愉快！

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慕运维9261880 提问者 #1

请问，如果run中循环调用加了synchronized关键字的函数，即使调用结束也不会切换线程，但是在循环中加上sleep就可以切换线程了，这是什么原因呢？

package thread;

public class test1 {
    int i = 0;
    int j = 0;
    boolean flag = false;

    public synchronized void get1() {
        System.out.println("1:" + ++i);
        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
//        System.out.println("get1 before wait");
//        if(flag) {
//            try {
//                wait();
//            } catch (InterruptedException e) {
//                e.printStackTrace();
//            }
//        }
//        System.out.println("get1 after wait");
//        flag = true;
//        notifyAll();
//        try {
//            Thread.sleep(5000);
//        } catch (InterruptedException e) {
//            e.printStackTrace();
//        }
    }

    public synchronized void get2() {
        System.out.println("2:" + ++j);
        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }


//        System.out.println("get2 before wait");
//        if(!flag) {
//            try {
//                wait();
//            } catch (InterruptedException e) {
//                e.printStackTrace();
//            }
//        }
//        System.out.println("get2 after wait");
//        flag = false;
//        notifyAll();
//        try {
//            Thread.sleep(100);
//        } catch (InterruptedException e) {
//            e.printStackTrace();
//        }
    }

}

package thread;

public class test1thread implements Runnable{
    test1 t1;
    test1thread(test1 t1){
        this.t1=t1;
    }
    @Override
    public void run() {
        while(true){
            t1.get1();
//            try {
//                Thread.sleep(1000);
//            } catch (InterruptedException e) {
//                e.printStackTrace();
//            }
        }
    }
}

package thread;

public class test1thread2 implements Runnable{
    test1 t1;
    test1thread2(test1 t1){
        this.t1=t1;
    }
    @Override
    public void run() {
        while(true){
            t1.get2();
//            try {
//                Thread.sleep(1000);
//            } catch (InterruptedException e) {
//                e.printStackTrace();
//            }
        }
    }
}