我有一个类似于此类,findbugz抱怨“从实例方法(initialize()和killStaticfield()写入静态字段”).我不能在ctor中设置静态字段.
>这个问题最好的解决办法是什么？
会把staticField放在AtomicReference中吗？
public class Something
{
private static SomeClass staticField = null;
private AnotherClass aClass;
public Something()
{
}
public void initialize()
{
//must be ctor'd in initialize
aClass = new AnotherClass();
staticField = new SomeClass( aClass );
}
public void killStaticField()
{
staticField = null;
}
public static void getStaticField()
{
return staticField;
}
}
解决方法
尽可能靠近您的原始设计…
public class Something {
private static volatile SomeClass staticField = null;
public Something() {
}
public static void getStaticField() {
if(Something.staticField == null)
Something.staticField = new SomeClass();;
return Something.staticField;
}
}
通过类名参考你的静态变量,这将删除findbugz警告.
将静态变量标记为volatile,这将使多参数环境中的引用更加安全.
更好的是：
public class Something {
private static final SomeClass staticField = new SomeClass();
public Something() {
}
public static void getStaticField() {
return Something.staticField;
}
}
