在c#中var关键字的一种用法是隐式类型声明。Java中var的等效语法是什么?
当前回答
你可以通过JetBrains查看Kotlin,但它是val。不是var。
其他回答
在Java 10中,等价的是……var。
没有。唉,你必须输入完整的类型名。
编辑:在发布7年后,Java 10中添加了局部变量的类型推断(使用var)。
编辑:在发布6年之后,收集下面的一些评论:
The reason C# has the var keyword is because it's possible to have Types that have no name in .NET. Eg: var myData = new { a = 1, b = "2" }; In this case, it would be impossible to give a proper type to myData. 6 years ago, this was impossible in Java (all Types had names, even if they were extremely verbose and unweildy). I do not know if this has changed in the mean time. var is not the same as dynamic. variables are still 100% statically typed. This will not compile: var myString = "foo"; myString = 3; var is also useful when the type is obvious from context. For example: var currentUser = User.GetCurrent(); I can say that in any code that I am responsible for, currentUser has a User or derived class in it. Obviously, if your implementation of User.GetCurrent return an int, then maybe this is a detriment to you. This has nothing to do with var, but if you have weird inheritance hierarchies where you shadow methods with other methods (eg new public void DoAThing()), don't forget that non-virtual methods are affected by the Type they are cast as. I can't imagine a real world scenario where this is indicative of good design, but this may not work as you expect: class Foo { public void Non() {} public virtual void Virt() {} } class Bar : Foo { public new void Non() {} public override void Virt() {} } class Baz { public static Foo GetFoo() { return new Bar(); } } var foo = Baz.GetFoo(); foo.Non(); // <- Foo.Non, not Bar.Non foo.Virt(); // <- Bar.Virt var bar = (Bar)foo; bar.Non(); // <- Bar.Non, not Foo.Non bar.Virt(); // <- Still Bar.Virt As indicated, virtual methods are not affected by this. No, there is no non-clumsy way to initialize a var without an actual variable. var foo1 = "bar"; //good var foo2; //bad, what type? var foo3 = null; //bad, null doesn't have a type var foo4 = default(var); //what? var foo5 = (object)null; //legal, but go home, you're drunk In this case, just do it the old fashioned way: object foo6;
一个简单的解决方案(假设你使用的是一个不错的IDE)是在所有地方输入'int',然后让它为你设置类型。
实际上我只是添加了一个名为var的类,所以我不需要输入不同的东西。
代码仍然过于冗长,但至少您不必键入它!
Java 10确实得到了局部变量类型推断,所以现在它有var,这与c#相当(据我所知)。
它还可以推断出不可标记的类型(程序员无法在该位置命名的类型;尽管哪些类型是不可标记的是不同的)。参见var和匿名类的技巧(你永远不应该在工作中使用)。
我能发现的一个区别是,在c#中,
如果一个名为var的类型在作用域中,那么var关键字将解析为该类型名称,并且不会被视为隐式类型局部变量声明的一部分。
在Java 10中,var不是一个合法的类型名称。
JEP - JDK增强提案
http://openjdk.java.net/jeps/286
JEP 286:局部变量类型推断
作者Brian Goetz
// Goals:
var list = new ArrayList<String>(); // infers ArrayList<String>
var stream = list.stream(); // infers Stream<String>
推荐文章
- 到底是什么导致了堆栈溢出错误?
- 为什么Android工作室说“等待调试器”如果我不调试?
- Java:路径vs文件
- ExecutorService,如何等待所有任务完成
- Maven依赖Servlet 3.0 API?
- 如何在IntelliJ IDEA中添加目录到应用程序运行概要文件中的类路径?
- getter和setter是糟糕的设计吗?相互矛盾的建议
- Android room persistent: AppDatabase_Impl不存在
- Java的String[]在Kotlin中等价于什么?
- Intellij IDEA上的System.out.println()快捷方式
- 使用Spring RestTemplate获取JSON对象列表
- Spring JPA选择特定的列
- URLEncoder不能翻译空格字符
- Java中的super()
- 如何转换JSON字符串映射<字符串,字符串>与杰克逊JSON