自定义枚举定义可能不是线程安全的。例如,
RoleEnum.java:
package com.threadsafe.bad;
public enum RoleEnum {
ADMIN(1),
DEV(2),
HEAD(3);
private Integer value;
private RoleEnum(Integer role){
this.value=role;
}
public static RoleEnum fromIntegerValue(Integer role){
for(RoleEnum x : values()){
if(x.value == role){
return x;
}
}
return RoleEnum.HEAD;
}
Class<?> buildFromClass;
public void setBuildFromClass(Class<?> classType){
buildFromClass=classType;
}
public Class<?> getBuildFromClass(){
return this.buildFromClass;
}
}
Main.java:
package com.threadsafe.bad;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
Thread threadA = new Thread(){
public void run(){
System.out.println("A started");
RoleEnum role;
role=RoleEnum.fromIntegerValue(1);
System.out.println("A called fromIntegerValue");
role.setBuildFromClass(String.class);
System.out.println("A called setBuildFromClass and start to sleep");
try {
Thread.sleep(10000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("Thread A: "+role.getBuildFromClass());
}
};
Thread threadB = new Thread(){
public void run(){
System.out.println("B started");
RoleEnum role;
role=RoleEnum.fromIntegerValue(1);
role.setBuildFromClass(Integer.class);
System.out.println("B called fromIntegerValue&setBuildFromClass and Start to sleep");
try {
Thread.sleep(20000);
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
System.out.println("B waked up!");
System.out.println("Thread B: "+ role.getBuildFromClass());
}
};
threadA.start();
threadB.start();
}
}
有时输出将是:
乙开始
b调用它fromIntegerValue &设置BuildFromClass并开始睡觉
一开始
一个名为fromIntegerValue
一个名为setBuildFromClass并开始睡觉
线程A:类java.lang.String
乙醒了!
线程B:类java.lang.String < - 我们期望java.lang。整数
有时输出将是:
一开始
一个名为fromIntegerValue
一个名为setBuildFromClass并开始睡觉
乙开始
b调用它fromIntegerValue & setBuildFromClass并开始睡觉
线程A:类java.lang.Integer < - 我们期待java.lang.String中
乙醒了!
线程B:类java.lang.Integer
你是什么意思的“线程安全”? 'leaveTheBuilding'方法不同步,所以当然可以同时运行多个线程。还是你在说初始化'INSTANCE'? – 2010-03-28 04:20:21
当你说“singleton”时,你的意思是它有可变状态吗?在这种情况下,无论如何你都会失败。 – 2010-03-28 04:46:15
只是我的$ 0.02,但我认为使用枚举来强制Singleton模式是代码混淆。也就是说,我知道布洛赫和其他很多人都认为这一点,我的评论可能是恐龙的咆哮。 – 2010-03-28 06:53:05