【2】GetSerialNumThread
package example.thread.threadLocal;
public class GetSerialNumThread implements Runnable {
public static void main(String args[]) {
GetSerialNumThread serialNumGetter = new GetSerialNumThread();
Thread t1 = new Thread(serialNumGetter, "Thread A");
Thread t2 = new Thread(serialNumGetter, "Thread B");
t1.start();
try {
t1.join();
} catch (InterruptedException e) {
e.printStackTrace();
}
t2.start();
}
public void run() {
int mySerialNum = getSerialNum();
System.out.println("线程 " + Thread.currentThread().getName() + " 获取到的序列号是" + mySerialNum); System.out.println("线程 " + Thread.currentThread().getName() + " 修改了序列号为" + (mySerialNum * 3));
setSerialNum(mySerialNum * 3);
System.out.println("线程 " + Thread.currentThread().getName() + " 再次获得的序列号是" + getSerialNum());
}
private int getSerialNum() {
return SerialNum.get();
}
private void setSerialNum(int newSerialNum) {
SerialNum.set(new Integer(newSerialNum));
}
}
运行的结果如下:
线程 Thread A 获取到的序列号是1
线程 Thread A 修改了序列号为3
线程 Thread A 再次获得的序列号是3
线程 Thread B 获取到的序列号是2
线程 Thread B 修改了序列号为6
线程 Thread B 再次获得的序列号是6
可见第一个线程在调用SerialNum.set(int)方法修改static变量时,其实修改的是它自己的副本,而不是修改本地变量,第二个线程在初始化的时候拿到的序列号是2而不是7。
为什么会这样呢?明明serialNum是静态变量啊?其实我们只需要看看ThreadLocal的内部构造就知道了:
A. ThreadLocal的get()方法:
/**
* Returns the value in the current thread\'s copy of this thread-local
* variable. Creates and initializes the copy if this is the first time
* the thread has called this method.
* * @return the current thread\'s value of this thread-local
*/
public T get() {
Thread t = Thread.currentThread();
ThreadLocalMap map = getMap(t);
if (map != null) return (T)
map.get(this); // Maps are constructed lazily. if the map for this thread // doesn\'t exist, create it, with this ThreadLocal and its // initial value as its only entry. T value = initialValue();
createMap(t, value);
return value;
}
可以看到ThreadLocal在内部维护了一个Map,将变量的值和线程绑定起来,get/set方法都是对该线程对应的value进行操作,所以不会影响到其它线程。