Spring和ThreadLocal

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在spring-framework的源码中,搜索了一下ThreadLocal,发现了如下结果,大家在阅读源码的时候,也可以使用这种方式,最起码,对一个类,一个方法的使用有更深的理解。

  • 地址:https://github.com/spring-projects/spring-framework/search?utf8=%E2%9C%93&q=ThreadLocal&type=Code

Spring和ThreadLocal_第1张图片

ThreadLocal到底有什么好处呢?

package com.lingdian.spring.test;


public class ThreadLocalTest implements Runnable{

	ThreadLocal num2 = new ThreadLocal();
	Integer num = 0;
	@Override
	public void run() {
		for(int i=0;i<10;i++){
			addNum2();
			addNum();
			System.out.println(Thread.currentThread().getName()+"[num:"+num+" num2:"+num2.get());
		}
	}
	void addNum2(){
		if(num2.get()==null){
			num2.set(0);
		}
		int m = num2.get().intValue();
		m++;
		num2.set(m);
	}
	void addNum(){
		 num++;
	}

	public static void main(String[] args) {
		ThreadLocalTest threadLocal1 = new ThreadLocalTest();
		new Thread(threadLocal1).start();
		new Thread(threadLocal1).start();
	}
	
}

打印结果:

Thread-1[num:2 num2:1
Thread-0[num:2 num2:1
Thread-0[num:4 num2:2
Thread-0[num:5 num2:3
Thread-1[num:3 num2:2
Thread-0[num:6 num2:4
Thread-0[num:8 num2:5
Thread-0[num:9 num2:6
Thread-0[num:10 num2:7
Thread-1[num:7 num2:3
Thread-0[num:11 num2:8
Thread-0[num:13 num2:9
Thread-0[num:14 num2:10
Thread-1[num:12 num2:4
Thread-1[num:15 num2:5
Thread-1[num:16 num2:6
Thread-1[num:17 num2:7
Thread-1[num:18 num2:8
Thread-1[num:19 num2:9
Thread-1[num:20 num2:10

从代码运行的结果可以看到变量num在Thread-0和Thread-1中被共享了。但是num2却没有。

转载于:https://my.oschina.net/lfy2008/blog/658149

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