Possible Duplicate: StringBuilder and StringBuffer in Java what is the difference between StringBuilder and Stringbuffer? Some methods in StringBuffer are synchronized while StringBuilder is not thread-safe - and faster. Rule of a thumb - use StringBuilder unless you have a use case, where a StringBuilder is used by more then one Thread (which would be a very rare case). Taken from the
可能重复： Java中的StringBuilder和StringBuffer StringBuilder和Stringbuffer有什么区别？ StringBuilder中的一些方法是同步的，而StringBuilder不是线程安全的 - 并且速度更快。 规则 - 使用StringBuilder，除非你有一个用例，其中一个StringBuilder被多个Thread使用（这将是一个非常罕见的情况）。 取自StringBuffer的javadoc： 从发行版JDK 5开始，此类已添加了一个旨在供单个线程{@link StringBuilder}使用的
This question already has an answer here: Difference between StringBuilder and StringBuffer 31 answers Local variables are thread-safe and variables declared inside the doFilter() method will be thread safe. Use StringBuilder for your purpose, as you shouldn't unnecessarily incur the overhead of synchronization used in StringBuffer . Moreover , The Servlet request and response objects
这个问题在这里已经有了答案： StringBuilder和StringBuffer之间的区别31个答案 局部变量是线程安全的，在doFilter()方法中声明的变量将是线程安全的。 为了您的目的使用StringBuilder ，因为您不应该不必要地招致在StringBuffer使用的同步开销。 而且，Servlet请求和响应对象是为每个新请求和响应重新创建的，因此它们本质上是线程安全的。 doFilter()方法将在每个请求的独立线程中执行。 推荐阅读： 为什么本地变
Possible Duplicate: StringBuilder and StringBuffer in Java Criteria to choose among StringBuffer and StringBuilder If you're definitely using Java 5 or higher, and you definitely don't need to share the object between threads (I can't remember ever needing to do so), StringBuilder is a better bet. Basically you should almost always use StringBuilder when you can, to avoid poin
可能重复： Java中的StringBuilder和StringBuffer 在StringBuffer和StringBuilder进行选择的标准 如果您肯定使用Java 5或更高版本，并且您绝对不需要在线程之间共享对象（我不记得曾经需要这么做），但StringBuilder是更好的选择。 基本上你应该尽可能使用StringBuilder ，以避免无意义的同步。 诚然，大多数虚拟机处理无争用同步的方式是非常有效的，但如果你不需要它，首先...... StringBuilder方法不同步，所以当
This question already has an answer here: Difference between StringBuilder and StringBuffer 31 answers Use StringBuilder class. It is more efficient at what you are trying to do. You should use the StringBuilder class. StringBuilder stringBuilder = new StringBuilder(); stringBuilder.append("Some text"); stringBuilder.append("Some text"); stringBuilder.append("Some text"); String finalSt
这个问题在这里已经有了答案： StringBuilder和StringBuffer之间的区别31个答案 使用StringBuilder类。 它对你所要做的事更有效率。 你应该使用StringBuilder类。 StringBuilder stringBuilder = new StringBuilder(); stringBuilder.append("Some text"); stringBuilder.append("Some text"); stringBuilder.append("Some text"); String finalString = stringBuilder.toString(); 另外，请访问： “Java如何用Stri
This question already has an answer here: How do I create a Java string from the contents of a file? 29 answers Yes, you can do this in one line (though for robust IOException handling you wouldn't want to). String content = new Scanner(new File("filename")).useDelimiter("\Z").next(); System.out.println(content); This uses a java.util.Scanner , telling it to delimit the input with Z ,
这个问题在这里已经有了答案： 如何从文件内容创建Java字符串？ 29个答案 是的，你可以在一行中做到这一点（尽管对于你不想要的健壮的IOException处理）。 String content = new Scanner(new File("filename")).useDelimiter("\Z").next(); System.out.println(content); 这使用一个java.util.Scanner ，告诉它用字符串锚点的结尾Z来分隔输入。 这最终使得输入有一个实际的令牌，这是整个文件，所以它可以通过一次调用ne
I'm using java.util.Scanner to read file contents from classpath with this code: String path1 = getClass().getResource("/myfile.html").getFile(); System.out.println(new File(path1).length()); // 22244 (correct) String file1 = new Scanner(new File(path1)).useDelimiter("\Z").next(); System.out.println(file1.length()); // 2048 (first 2k only) Code runs from idea with command (maven test) /L
我使用java.util.Scanner从以下代码的classpath中读取文件内容： String path1 = getClass().getResource("/myfile.html").getFile(); System.out.println(new File(path1).length()); // 22244 (correct) String file1 = new Scanner(new File(path1)).useDelimiter("\Z").next(); System.out.println(file1.length()); // 2048 (first 2k only) 代码从想法和命令运行（maven测试） /Library/Java/JavaVirtualMachines/jdk1
I am trying to set up Hibernate. But when i try to create my Session Factory, with this code: Configuration configuration = new Configuration(); configuration.configure(); serviceRegistry = new ServiceRegistryBuilder().applySettings(configuration.getProperties()).buildServiceRegistry(); sessionFactory = configuration.buildSessionFactory(serviceRegistry); I get the error: org.hibernate
我正在尝试设置Hibernate。 但是当我尝试创建我的Session Factory时，使用以下代码： Configuration configuration = new Configuration(); configuration.configure(); serviceRegistry = new ServiceRegistryBuilder().applySettings(configuration.getProperties()).buildServiceRegistry(); sessionFactory = configuration.buildSessionFactory(serviceRegistry); 我收到错误： org.hibernate.cfg.beanvalidatio
Programming in Android, most of the text values are expected in CharSequence . Why is that? What is the benefit, and what are the main impacts of using CharSequence over String? What are the main differences, and what issues are expected, while using them, and converting from one to another? Strings are CharSequences, so you can just use Strings and not worry. Android is merely trying to
在Android中编程时，大部分文本值都在CharSequence 。 这是为什么？ 什么是好处，以及使用CharSequence不是String的主要影响是什么？ 主要的区别是什么？在使用它们时将会遇到哪些问题？从一个转换到另一个？ 字符串是CharSequences，所以你可以使用Strings而不用担心。 Android只是试图通过允许您指定其他CharSequence对象（如StringBuffers）来帮助您。 我相信最好使用CharSequence。 原因是String实现了CharSequen
What happens (behind the curtains) when this is executed? int x = 7; x = x++; That is, when a variable is post incremented and assigned to itself in one statement? I compiled and executed this. x is still 7 even after the entire statement. In my book, it says that x is incremented! x does get incremented. But you are assigning the old value of x back into itself. x = x++; x++ increments
执行此操作后会发生什么（幕后）？ int x = 7; x = x++; 也就是说，当一个变量在一个语句中增加后分配给自己？ 我编译并执行了这个。 即使在整个陈述之后， x仍然是7。 在我的书中，它说x增加了！ x确实增加了。 但是你将x的旧值重新分配给自己。 x = x++; x++递增x并返回其旧值。 x =将旧值重新分配给自己。 所以最后， x被赋值回它的初始值。 x = x++; 相当于int tmp = x; x++; x = tmp; 该声明： x = x++;
The method signature of a Java main() method is: public static void main(String[] args){ ... } Is there a reason for this method to be static? The method is static because otherwise there would be ambiguity: which constructor should be called? Especially if your class looks like this: public class JavaClass{ protected JavaClass(int x){} public void main(String[] args){ } } Should
Java main（）方法的方法签名是： public static void main(String[] args){ ... } 这种方法是否有静态的原因？ 这个方法是静态的，否则就会产生歧义：应该调用哪个构造函数？ 特别是如果你的班级看起来像这样： public class JavaClass{ protected JavaClass(int x){} public void main(String[] args){ } } JVM应该调用new JavaClass(int)吗？ 它应该通过什么样的x ？ 如果不是，JVM是否应该实例化JavaCla