JAVA多线程编程核心技术--遇到的问题
sleep(1)可以解决共享内存的问题
代码
- 一个线程B写,一个线程A读,在没有volitile的情况下,如何停止线程A。
ThreadB.java
package com;
public class ThreadB extends Thread {
private MyList list;
public ThreadB(MyList list) {
this.list = list;
}
@Override
public void run() {
try {
for (int i = 0; i < 10; i++) {
list.add();
System.out.println("添加了" + i + "个元素");
Thread.sleep(1000);
}
} catch (Exception e) {
e.printStackTrace();
}
}
}
package com;
public class ThreadA extends Thread {
private MyList list;
public ThreadA(MyList list) {
this.list = list;
}
@Override
public void run() {
try {
while (true) {
sleep(1);
if (list.size() == 5) {
System.out.println("线程退出");
throw new InterruptedException();
}
}
} catch (Exception e) {
System.out.println("异常了");
e.printStackTrace();
}
}
}
package com;
import java.util.ArrayList;
import java.util.List;
public class MyList {
private List<String> list = new ArrayList<String>();
public void add() {
list.add("毕玉强");
}
public int size() {
return list.size();
}
}
package com;
public class Run {
public static void main(String[] args) {
MyList myList = new MyList();
ThreadA threadA = new ThreadA(myList);
threadA.setName("a");
ThreadB threadB = new ThreadB(myList);
threadB.setName("b");
threadB.start();
threadA.start();
}
}
问题解释
- 虽然没有volitile,sleep(1)会引起上下文切换,将主内存的数据重新加载到线程栈。
- 另外jvm会尽可能的将数据
加载
,到主内存中,并且在主内存读读取
。
- System.out.PrintLn(" "),sychronized(list){},因为有个this锁,所以比较耗时,也符合上述条件。