[英]volatile keyword : Need explanation for this program behaviour
通過此程序了解volatile關鍵字:
public class VolatileExample implements Runnable{
private volatile int vol;
@Override
public void run(){
vol=5;
while(vol==5)
{
System.out.println("Inside run when vol is 5. Thread is : "+Thread.currentThread().getName());
if("t2".equals(Thread.currentThread().getName()))
{
System.out.println("Got Thread : "+Thread.currentThread().getName()+" Now Calling Stop To End This Flow");
stopIt();
}
}
}
public void stopIt(){
vol=10;
}
public static void main(String[] args){
VolatileExample ve1 = new VolatileExample();
VolatileExample ve2 = new VolatileExample();
Thread t1 = new Thread(ve1);
Thread t2 = new Thread(ve1); //t1 and t2 operate on same instance of VolatileExample class
t1.setName("t1");
t2.setName("t2");
t1.start();
t2.start();
}
}
輸出:
Inside run when vol is 5. Thread is : t1
Inside run when vol is 5. Thread is : t1
Inside run when vol is 5. Thread is : t2
Inside run when vol is 5. Thread is : t1
Got Thread : t2 Now Calling Stop To End This Flow
Inside run when vol is 5. Thread is : t1
所有對vol變量的寫入將立即寫入主內存,並且應該立即對其他線程“可見”。 為什么在調用stopIt()之后t1線程仍然執行? 想知道現在的vol值是10而不是5嗎?
沒有證據表明在調用stopIt()
之后t1
正在運行。
事情可能以這種順序發生
t1 t2
System.out.println("calling stopIt()");
while(vol==5)
System.out.println("Inside run")
enter stopIt()
vol = 10
這會給您觀察到的結果。 (還有其他可能的訂購結果,您可以得到此結果。)
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