I have an XML file containing many tags. Out of those some can be missing. So while reading such unmarshalled file using JAXB I am getting NullPointerException for those missing tags. I know the one solution that I can check whether object is null or not. But it's not convenient to check null condition for each and every tag because there are too many tags. If I handle NullPointerExcept
我有一个包含许多标签的XML文件。 有些人可能会失踪。 所以，当使用JAXB读取这样的非编组文件时，我得到了那些缺少标签的NullPointerException。 我知道我可以检查对象是否为null的一个解决方案。 但检查每个标签的空条件并不方便，因为标签太多。 如果我处理NullPointerException，它只会处理异常，但不会进一步读取xml。 有没有任何有效的方法来处理这种情况（特别是如何摆脱检查每个标签的空值？） Request rqstr = b
I wonder if it would be possible to 'avoid' null checks in Java, take an example from this code: @Override public List<AccountBean> search(AccountConstraint... c) { if (c.length == 0) { throw new IllegalArgumentException("dao.AccountDAO.search: c.length == 0"); } try { List<AccountBean> beans = new ArrayList<>(); for (AccountConstrain
我想知道是否可以'避免'在Java中的空检查，从这个代码的例子： @Override public List<AccountBean> search(AccountConstraint... c) { if (c.length == 0) { throw new IllegalArgumentException("dao.AccountDAO.search: c.length == 0"); } try { List<AccountBean> beans = new ArrayList<>(); for (AccountConstraint ac : c) { Builder builde
A web service returns a huge XML and I need to access deeply nested fields of it. For example: return wsObject.getFoo().getBar().getBaz().getInt() The problem is that getFoo() , getBar() , getBaz() may all return null . However, if I check for null in all cases, the code becomes very verbose and hard to read. Moreover, I may miss the checks for some of the fields. if (wsObject.getFoo() ==
Web服务返回一个巨大的XML，我需要访问它的深层嵌套字段。 例如： return wsObject.getFoo().getBar().getBaz().getInt() 问题是getFoo() ， getBar() ， getBaz()可能全都返回null 。 但是，如果我在所有情况下都检查为null ，则代码会变得非常冗长，难以阅读。 而且，我可能会错过对某些领域的检查。 if (wsObject.getFoo() == null) return -1; if (wsObject.getFoo().getBar() == null) return -1; // maybe also do s
This question already has an answer here: Is synchronized inherited in Java? 3 answers If a class has synchronized methods, does its subclass also have the same synchronized methods, whether simply inherited or overridden by the subclass? A synchronized method from a super class can be overriden in a subclass as a non synchronized method and vice-versa. If the subclass does not override
这个问题在这里已经有了答案： 在Java中是否同步继承？ 3个答案 如果一个类有同步方法，它的子类是否也具有相同的同步方法，无论是简单地继承还是被子类重写？ 来自超类的synchronized方法可以在子类中被重写为非synchronized方法，反之亦然。 如果子类不覆盖synchronized方法但只是继承它，那么该方法将作为synchronized方法按原样继承。 我想知道我们是否需要找到堆栈和属性的同步替换 Stack和Properties是线程安
I have a question which is related to the Stack implementation in the Collections framework in Java. I can see from the implementation that the size of the Stack can grow. Does this mean that a StackOverflowError can never occur and eventually the Stack reaches a size and an OutOfMemoryError occurs? From googling I found that the Vector class is deprecated since it synchronizes each and ever
我有一个问题，它与Java中的Collections框架中的Stack实现有关。 从实现中我可以看到堆栈的大小可以增长。 这是否意味着StackOverflowError永远不会发生，并最终堆栈达到一个大小，并发生OutOfMemoryError ？ 从Google搜索中我发现Vector类已被弃用，因为它像Jon Skeet在这里指出的那样同步每一个操作：是不是弃用了Java向量？ 那么，在这之后，我会使用这个Java类的真实场景吗？ 我不想在每一个操作上同步，并且想要在
This question already has an answer here: Are fields initialized before constructor code is run in Java? 4 answers Instance variable initialization expressions such as int var = getVal(); are evaluated after the super class constructor is executed but prior to the execution of the current class constructor's body. Therefore getVal() is called before the body of the ChkCons constructor
这个问题在这里已经有了答案： 在Java中运行构造函数代码之前，字段是否已初始化？ 4个答案 实例变量初始化表达式，如int var = getVal(); 在超类构造函数执行之后但在执行当前类构造函数的主体之前进行求值。 因此，在执行ChkCons构造函数的主体之前调用getVal() 。 构造函数在方法之前调用。 方法的执行发生在作为对象实例变量评估对象创建的一部分之后。 从以下代码可以更好地理解这一点。 class SuperClass{
为什么你不能在Java中声明一个类是静态的？ Only nested classes can be static. By doing so you can use the nested class without having an instance of the outer class. class OuterClass{ public static class StaticNestedClass{ } public class InnerClass{ } public InnerClass getAnInnerClass(){ return new InnerClass(); } //This method doesn't work public static
为什么你不能在Java中声明一个类是静态的？ 只有嵌套的类可以是静态的。 通过这样做，您可以使用嵌套类而不需要外部类的实例。 class OuterClass{ public static class StaticNestedClass{ } public class InnerClass{ } public InnerClass getAnInnerClass(){ return new InnerClass(); } //This method doesn't work public static InnerClass getAnInnerClassStatically(){
Waaah, the Play! framework has so many static methods. Where I go to school, we were told never ever to use any statics, yet Play! uses it like there's no tomorrow. Is that somehow okay? If so, why? We (7 people and I) are planning to use the Play! framework for a project involving a web app. We decided to do it with Play! because it looks quite fun to do, all of us already know Jav
Waaah，玩！ 框架有很多静态方法。 我在哪里上学，我们被告知永远不会使用任何静力学，但玩！ 像没有明天一样使用它。 那好吗？ 如果是这样，为什么？ 我们（7人和我）计划使用Play！ 涉及Web应用程序的项目框架。 我们决定用Play来完成！ 因为它看起来相当有趣，我们所有人都已经知道Java并且任务非常艰难，所以我们想要专注于实际的任务，而不是学习如何用不同的语言编程。 然而，我们总是被告知，在我们开发的任
This question already has an answer here: In what situations is the CopyOnWriteArrayList suitable? [duplicate] 2 answers Based on the javadoc, A thread-safe variant of java.util.ArrayList in which all mutative operations (add, set, and so on) are implemented by making a fresh copy of the underlying array. As you can see, any modification to the container is costly especially when the siz
这个问题在这里已经有了答案： 在什么情况下CopyOnWriteArrayList适合？ [重复] 2个答案 基于javadoc， java.util.ArrayList的线程安全变体，其中所有可变操作（add，set等）都通过创建底层数组的全新副本来实现。 正如你所看到的，对容器的任何修改都是昂贵的，特别是当容器中的数据量很大时。 当修改频率较低时，最好使用CopyOnWriteArrayList 。 正如JCIP所建议的那样 ， 写入时复制集合仅适用于迭代比修改更普
I have following case, public class Test { private static final int MAX_NUMBER = 10_00_00; public static void main(String[] args) { List<Integer> list = new CopyOnWriteArrayList<>(); long start = System.nanoTime(); for(int i = 0; i < MAX_NUMBER; i++) { list.add(i * 2); } long end = System.nanoTime(); System.ou
我有以下情况， public class Test { private static final int MAX_NUMBER = 10_00_00; public static void main(String[] args) { List<Integer> list = new CopyOnWriteArrayList<>(); long start = System.nanoTime(); for(int i = 0; i < MAX_NUMBER; i++) { list.add(i * 2); } long end = System.nanoTime(); System.out.printl