synchronized 线程同步

synchronized  通常用来形容一次方法的调用,调用一旦开始,调用者必须等到方法调用返回后,才能继续执行后续的操作。

1.demo

package demo1;
public class MyThread extends Thread {
    private  int count = 5;
    @Override
    public  void run() {
        count--;
        System.err.println("∽"+currentThread().getName()+"count:"+this.count);
    }
    public  static void main(String[] args){
        MyThread  myThread = new MyThread();
        Thread t1 = new Thread(myThread,"myThread-1");
        Thread t2 = new Thread(myThread,"myThread-2");
        Thread t3 = new Thread(myThread,"myThread-3");
        Thread t4 = new Thread(myThread,"myThread-4");
        Thread t5 = new Thread(myThread,"myThread-5");
        t1.start();
        t2.start();
        t3.start();
        t4.start();
        t5.start();

    }
}
结果输出:

        ∽myThread-1count:2
        ∽myThread-4count:1
        ∽myThread-2count:2
        ∽myThread-3count:2
        ∽myThread-5count:0

 

描述:5个线程都会去对这个 count  变量进行操作,但是出现了重复的数据,说明没有进行同步、共享。

2.demo:使用 synchronized 关键字

package demo1;

public class MyThread extends Thread {


    private  int count = 5;
    @Override
    public synchronized void run() {
        count--;
        System.err.println("∽"+currentThread().getName()+"count:"+this.count);
    }
    public  static void main(String[] args){
        MyThread  myThread = new MyThread();
        Thread t1 = new Thread(myThread,"myThread-1");
        Thread t2 = new Thread(myThread,"myThread-2");
        Thread t3 = new Thread(myThread,"myThread-3");
        Thread t4 = new Thread(myThread,"myThread-4");
        Thread t5 = new Thread(myThread,"myThread-5");
        t1.start();
        t2.start();
        t3.start();
        t4.start();
        t5.start();

    }
}

结果输出:

      ∽myThread-1count:4
      ∽myThread-3count:3
      ∽myThread-2count:2
      ∽myThread-4count:1
      ∽myThread-5count:0

 

描述:输出的结果没有重复的,说明同步生效了。

  1.当多个线程都去访问 MyThread 的run 方法的时候,都会以队列的的方式去进行访问处理(CPU的分配)。

  2.一个线程要执行 synchronized 方法修饰的代码块:都会去尝试去获取锁(object的锁),如果拿到锁,则执行 synchronized 代码的内容;

   如果拿不到,则会不断的去尝试获取这把锁,一直到拿到为止 (多个线程去尝试获取这把锁,就行形成竞争、排队的问题)。

原文地址:https://www.cnblogs.com/xxt19970908/p/6954157.html