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>

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;

如果您将Lombok添加到项目中，您可以使用它的val关键字。

http://projectlombok.org/features/val.html

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>

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!

在Java 10中，等价的是……var。