Scala替换Arrays.binarySearch？ [英] Scala replacement for Arrays.binarySearch?

问题描述

在Scala for Java的 int Arrays.binarySearch（Object []数组，对象）中是否有替换？

<问题是Scala的数组不是协变的，所以我必须首先投射我的 stringArray：Array [String] ：

  stringArray.asInstanceOf [Array [Object]] 
  

有更好的解决方案吗？

解决方案

据我所知，目前还没有内置任何东西，但是你可以使用 pimp-my-library模式很容易实现这一目标。像这样：

  class ObjectArrayTools [T< ;: AnyRef]（a：Array [T]）{
 def binarySearch（key：T）= {
 java.util.Arrays.binarySearch（a.asInstanceOf [Array [AnyRef]]，key）
} 
} 
隐式def anyrefarray_tools [ T< ;: AnyRef]（a：Array [T]）= new ObjectArrayTools（a）
 
 scala>数组（a，fish，is，some，thing）。binarySearch（some）
 res26：Int = 3 
 scala>数组（a，fish，is，some，thing）。binarySearch（bye）
 res28：Int = -2 
  

如果需要，可以将其他 java.util.Arrays 对象方法添加到同一个类中它们也是。

一般来说，我觉得习惯于总是导入你最喜欢的Scala实用程序的集合是个好主意。添加这样的功能非常容易，你可以这样做，而不是继续输入 .asInstanceOf [Array [AnyRef]] ，只需要一点努力就可以了让自己显着提高工作效率。

Is there a replacement in Scala for Java's int Arrays.binarySearch(Object[] array, object)?

The problem is that Scala's Arrays are not covariant, so I would have to cast my stringArray: Array[String] like this first:

stringArray.asInstanceOf[Array[Object]]

Is there a better solution?

解决方案

There isn't anything built in as far as I know, but you can use the pimp-my-library pattern to accomplish this fairly easily. Like so:

class ObjectArrayTools[T <: AnyRef](a: Array[T]) {                  
   def binarySearch(key: T) = {
     java.util.Arrays.binarySearch(a.asInstanceOf[Array[AnyRef]],key)
   }
}
implicit def anyrefarray_tools[T <: AnyRef](a: Array[T]) = new ObjectArrayTools(a)

scala> Array("a","fish","is","some","thing").binarySearch("some")
res26: Int = 3
scala> Array("a","fish","is","some","thing").binarySearch("bye")  
res28: Int = -2

You can add the other java.util.Arrays object methods into the same class if you need them too.

In general, I find it a good idea to get used to always importing a collection of your favorite Scala utilities. It's so easy to add functionality like this that you may as well do it in general rather than keep typing .asInstanceOf[Array[AnyRef]], and with a little effort you can make yourself significantly more productive.

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