In Swift, function types are treated as first class citizens, and can be used anywhere, as any other types. Its syntax is straightforward, and can be easily understood and used. On the other hand, Java does support function programming via lambda expressions, but in general, I have found it less intuitive, harder to understand, and seemingly more restrictive, compared to Swift's function t
在Swift中，函数类型被视为第一类公民，并且可以在任何地方使用，就像任何其他类型一样。 它的语法很简单，可以很容易理解和使用。 另一方面，Java不支持通过lambda表达式进行函数编程，但总的来说，与Swift的函数类型相比，我发现它不那么直观，难以理解，并且看起来更具限制性。 我的问题是，既然Swift的函数类型和Java的lambda表达式的目的是实现函数式编程，那么Swift的函数类型可以做什么，但Java的lambda表达式不能？
When we learn C++ in school, our professor will tell us to write return 0; at the last line of codes in the main function and it is considered as a good programming practice. In Java, I realise some people writes System.exit(0); at the last line of the main method. However, in C++, if I use exit(0); I get penalized by my professor, because (in school) for procedural programming, we are exp
当我们在学校学习C ++时，我们的教授会告诉我们写return 0; 在主函数的最后一行代码中，它被认为是一种很好的编程习惯。 在Java中，我意识到一些人写了System.exit(0); 在主要方法的最后一行。 但是，在C ++中，如果我使用exit(0); 我受到我的教授的惩罚，因为（在学校）进行程序编程，我们希望让程序一直到主要结束，让程序自然停止。 我的问题是：Java的System.exit(0); 类似于C ++的return 0; ？ （或者它类似于C
I have a situation where i have to access a non static member from inside a static method. I can access it with new instance, but current state will be lost as non static member will be re-initialized. How to achieve this without losing data? Maybe you want a singleton. Then you could get the (only) instance of the class from within a static method and access its members. The basic idea is
我有一种情况，我必须从静态方法中访问非静态成员。 我可以用新实例访问它，但是当非静态成员将被重新初始化时，当前状态将会丢失。 如何实现这一点而不会丢失数据？ 也许你想要一个单身人士。 然后，您可以从静态方法中获取该类的（唯一）实例并访问其成员。 基本的想法是 public class Singleton { private static Singleton instance = null; private Singleton() {} public static Singleton getInstance() {
When is it considered poor practice to use the static keyword in Java on method signatures? If a method performs a function based upon some arguments, and does not require access to fields that are not static, then wouldn't you always want these types of methods to be static? One reason why you may not want it to be static is to allow it to be overridden in a subclass. In other words, the
什么时候在方法签名中使用Java中的static关键字被认为是不好的做法？ 如果一个方法根据一些参数执行一个函数，并且不需要访问非静态的字段，那么你是否总是希望这些类型的方法是静态的？ 你可能不希望它是静态的一个原因是允许它在一个子类中被覆盖。 换句话说，行为可能不依赖于对象内的数据，而取决于对象的确切类型。 例如，你可能有一个通用的集合类型，与isReadOnly财产将返回false的永远可变集合， true的永远不变的
I'm trying to use lambda expressions and streams in my Android project.I'm using streamsupport library for streams, and native Java 8 for lambda expressions. To be able to use Java 8 features I need to add compileOptions { sourceCompatibility JavaVersion.VERSION_1_8 targetCompatibility JavaVersion.VERSION_1_8 } this code in my gradle, but the compiler shows this error: Error:J
我试图在我的Android项目中使用lambda表达式和流。我使用流支持库来处理流，并使用原生Java 8处理lambda表达式。 为了能够使用我需要添加的Java 8功能 compileOptions { sourceCompatibility JavaVersion.VERSION_1_8 targetCompatibility JavaVersion.VERSION_1_8 } 这个代码在我的gradle中，但是编译器显示这个错误： Error:Jack is required to support java 8 language features. Either enable Jack or remove s
I am using IntelliJ IDEA version 2017.3, then I install plug-in CheckStyle-IDEA v 5.16.0 successful, I install Apache Maven Checkstyle Plugin to run myCheckFile.xml . At install phase, I received many check style error like: Indentation: '.' has incorrect indentation level 6, expected level should be 8. Indentation: 'new' has incorrect indentation level 6, expected level should be 8. After imp
我使用IntelliJ IDEA版本2017.3，然后安装插件CheckStyle-IDEA v 5.16.0成功，我安装Apache Maven Checkstyle插件以运行myCheckFile.xml 。 在install阶段，我收到了许多检查样式错误，如： Indentation: '.' has incorrect indentation level 6, expected level should be 8. Indentation: 'new' has incorrect indentation level 6, expected level should be 8. 导入CheckStyle设置后，我期望当我在IntelliJ IDEA中按Ctrl
I am trying to import the following checkstyle file into the CheckStyle-IDE plugin which I already installed on IntelliJ but it doesn't let me do that ( it gives me errors) ONOS checkstyle file I can import it as a CodeStyle config file (Settings-->Editor-->CodeStyle) but not as a CheckStyle config file. When I run mvn clean install on my project, it gives me checkStyle violations
我试图将下面的checkstyle文件导入到我已经安装在IntelliJ上的CheckStyle-IDE插件中，但它不让我这样做（它给我错误） ONOS checkstyle文件 我可以将它作为CodeStyle配置文件（设置 - >编辑 - > CodeStyle）导入，但不能作为CheckStyle配置文件导入。 当我在我的项目上运行mvn clean install时，它会给我checkStyle违规，但我看不到它们。 我如何使用CheckStyle插件查看mvn checkstyle违规？ 谢谢
I have maven project with maven-checkstyle-plugin. After maven build I want to see all checkstyle errors (may be in html page). I install Sonar plugin to Idea, but what next? I can not any window or item in menu, I run mvn checkstyle:check and nothing happened in GUI. So is it possible to see all checkstyle errors (not the xml file in target folder) with Intellij Idea? Why don't you use
我有maven-checkstyle-plugin的maven项目。 maven构建后，我想看到所有的checkstyle错误（可能在html页面中）。 我将Sonar插件安装到Idea上，但接下来要做什么？ 我不能在菜单中的任何窗口或项目，我运行mvn checkstyle：检查和GUI中没有发生任何事情。 那么是否可以使用Intellij Idea查看所有checkstyle错误（不是目标文件夹中的xml文件）？ 你为什么不使用Idea checkstyle插件 - http://plugins.jetbrains.com/plugin?p
I have a small question about IntelliJ IDEA 11. I just imported a project from subversion - its a maven project. But I have a problem in maven library dependencies so that I can't include all maven dependencies automatically - IDEA shows dependency errors only when I open that class/ Thats what I get here: So I want all dependencies to be added automatically - is that possible or do I have
我有一个关于IntelliJ IDEA 11的小问题。我刚刚从subversion导入了一个项目 - 它是一个maven项目。 但是我在Maven库依赖关系中遇到了一个问题，所以我无法自动包含所有Maven依赖关系 - IDEA仅在打开该类时才显示依赖关系错误/这就是我在这里得到的结果： 所以我想要所有的依赖项自动添加 - 这是可能的，还是我必须通过所有类文件来识别和添加Maven依赖关系？ 更新：做了一些修改后，我发现如何以某种方式解决我的问题。
I was hoping someone would be able to help me out. I am making a Sudoku game, and I wrote a class which solves the entire puzzle (and it is of course connected to an action button in another java file), and it works well. What I wanted to know is if someone could point me in the right direction if I wanted to make a "hint" button, which randomly applies the correct value to just one
我希望有人能够帮助我。 我正在制作一款Sudoku游戏，并且我编写了一个解决整个难题的类（它当然与另一个java文件中的动作按钮相关），并且它运行良好。 我想知道的是，如果我想制作一个“提示”按钮，如果有人能够指引我正确的方向，它会随机将正确的值应用于一个单独的方块。 这是求解器代码： public class SudokuSolver { // set unchangeable number of rows to 9 private static final int ROWS = 9; // set unchangeab