Scala：如何强制将整数包装为对象？ [英] Scala: How to force wrapping an integer as an object?

问题描述

我在这里收到错误：

val a: Int = 1
val i: Int with Object = a

如何将此1转换为scala中的整数对象？
我的目的是将它传递给数组[带对象的Int] 。
目前显示错误：

How can I convert this 1 to an integer object in scala? My purpose is to pass it to an Array[Int with Object]. It currently displays the error:

error type mismatch
found : Int(1)
required: Int with java.lang.Object
       val i: Int with Object = a
                                ^

编辑

我有这个错误，因为我使用的是android ArrayAdapter ，以及因此，通过定义：

I have this error because I am using an android ArrayAdapter from scala, and therefore by defining:

class ImageAdapter[T](ctx: Context, viewResourceId: Int, pointers: Array[T]) extends ArrayAdapter[T](ctx, viewResourceId, pointers) { ... }

它会抛出这个错误：

overloaded method constructor ArrayAdapter with alternatives: 
(android.content.Context,Int,java.util.List[T])android.widget.ArrayAdapter[T] <and> 
(android.content.Context,Int,Array[T with Object])android.widget.ArrayAdapter[T] <and> 
(android.content.Context,Int,Int)android.widget.ArrayAdapter[T]
   cannot be applied to (android.content.Context,  Int, Array[T])

所以我需要用替换 T 类中的T<：Object 类ImageAdapter [T< ;: Object]（ctx：...

So I need to replace T with T <: Object in class ImageAdapter[T <: Object](ctx: ...

推荐答案

Int 是一种scala类型，通常映射到java的 int ，但是在装箱时会映射到 java.lang.Integer 。无论是否装箱，scala都是透明的。

Int is a scala type which usually maps to java's int, but will map to java.lang.Integer when boxed. Whether it's boxed or not is mostly transparent in scala.

在任何情况下， Int 绝对不是 java.lang.Object 的子类型。实际上 Int 是 AnyVal 的子类型，它不是的子类型java.lang.Object 。因此，带对象的I​​nt是非常荒谬的，因为你不能拥有任何一个<$ c的具体类型$ c> Int 和一个 java.lang.Object 。

In any case, Int is definitely not a sub-type of java.lang.Object. In fact Int is a sub-type of AnyVal which is not a sub-type of java.lang.Object. Thus Int with Object is pretty much nonsensical, given that you cannot have any concrete type that is both an Int and a java.lang.Object.

我想是什么您意思是这样的：

I think what you meant is rather something like:

val i: Object = a

或者更多的是：

val i: AnyRef = a

当然，这些都没有编译，但你可以强制装箱 Int 值转换为 AnyRef ：

Of course, none of this compiles, but you can force the boxing of the Int value by casting to AnyRef:

val i: AnyRef = a.asInstanceOf[AnyRef]

与一般情况不同，施放 AnyVal 到 AnyRef 总是安全的，并会强制装箱。

Unlike in the general case, casting an AnyVal to an AnyRef is always safe, and will force the boxing.

您还可以使用更具体的 Int.box 函数：

You can also use the more specific Int.box function:

val i: AnyRef = Int.box(a)

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