使用泛型获取列表类：例如：List< Number> :: class [英] Getting Class of list with generic eg: List<Number>::class

问题描述

我有一个一般类型的类 Builder< T> ，它带有一个构造函数参数 Class< T> 可以保持周围的类型。这是一个我在java代码中使用很多的类，所以我不想更改签名。
当我尝试使用这样的构造函数时：
$ b

Builder< List< Number>>（List< Number> :: class）

我得到一个错误：类类只允许在类文字的左边

有什么办法解决这个问题？
我无法更改 Builder 的构造函数，太多的java类依赖它。

我明白整个类型的擦除问题，我真的只是想让编译器很开心。 解决方案

由于泛型类型擦除 List 类只有一个实现所有的通用实例。你只能得到一个对应于 List<> 类型的类，因此只能创建 Builder< List<>> 。

该构建器实例适合构建 列表。再次由于类型擦除，你可以通过未经检查的强制转换来自行决定： ：class.java）as Builder< List< Number>>
Builder（List :: class.java as Class< List< Number>>）

另一种方法是创建内联通用辅助函数：

 内联fun   

并以如下方式使用它：

  Builder< List<>>（）
  

I have a generically typed class Builder<T> that takes a constructor argument Class<T> so I can keep the type around. This is a class that I use a lot in java code so I don't want to change the signature. When I try to use the constructor like this:

Builder<List<Number>>(List<Number>::class)

I get an error: "Only classes are allowed on the left hand side of a class literal"

Any way to resolve this? I can't change the constructor for Builder, too many java classes rely upon it.

I understand the whole type erasure issue, I really just want to make the compiler happy.

解决方案

Due to generic type erasure List class has a single implementation for all its generic instantiations. You can only get a class corresponding to List<*> type, and thus create only Builder<List<*>>.

That builder instance is suitable for building a list of something. And again due to type erasure what that something is you can decide by yourself with the help of unchecked casts:

Builder(List::class.java) as Builder<List<Number>>
Builder(List::class.java as Class<List<Number>>)

Another approach is to create inline reified helper function:

inline fun <reified T : Any> Builder() = Builder(T::class.java)

and use it the following way:

Builder<List<Number>>()

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