[英]How to convert Scala List to Java ArrayList
I am using a Java API that requires ArrayList as parameter. 我使用的是需要ArrayList作为参数的Java API。 Now in Scala, I have a List[String].
现在在Scala中,我有一个List [String]。 How can I convert the List[String] to ArrayList in Scala?
如何在Scala中将List [String]转换为ArrayList?
I have tried : 我试过了 :
import scala.collection.JavaConverters._
val scalaList = List("a","b","c")
scalaList.asJava
Result is: 结果是:
java.util.List[String] = [a, b, c]
The above does not work because I want the result to be 以上不起作用,因为我想要结果
java.util.ArrayList[String]
的java.util.ArrayList [字符串]
instead of 代替
java.util.List[String] = [a, b, c]
java.util.List [String] = [a,b,c]
It totally makes sense to return java.util.List[T]
which is a java interface and can have ArrayList[T]
or LinkedList[T]
as implementation. 返回
java.util.List[T]
是完全有意义的,它是一个java接口,可以将ArrayList[T]
或LinkedList[T]
作为实现。
scala> val scalaList = List(1,2,3)
scalaList: List[Int] = List(1, 2, 3)
scala> import scala.collection.JavaConverters._
import scala.collection.JavaConverters._
scala> scalaList.asJava
res22: java.util.List[Int] = [1, 2, 3]
In my opinion it needs to be converted to ArrayList[T]
or LinkedList[T]
whatever you want by yourself. 在我看来,无论你想要什么,它都需要转换为
ArrayList[T]
或LinkedList[T]
。
One way I could think of is 我能想到的一种方法是
scala> new java.util.ArrayList[Int](scalaList.asJava)
res27: java.util.ArrayList[Int] = [1, 2, 3]
or 要么
scala> new java.util.LinkedList[Int](scalaList.asJava)
res28: java.util.LinkedList[Int] = [1, 2, 3]
Better version of this would be 更好的版本是
scala> def toArrayList[T](input: List[T]): java.util.ArrayList[T] = new java.util.ArrayList[T](input.asJava)
toArrayList: [T](input: List[T])java.util.ArrayList[T]
scala> toArrayList(List(1000, 2000, 3000))
res33: java.util.ArrayList[Int] = [1000, 2000, 3000]
And the best version would be, write implicit method 最好的版本是,写隐式方法
class AsArrayList[T](input: List[T]) {
def asArrayList : java.util.ArrayList[T] = new java.util.ArrayList[T](input.asJava)
}
implicit def asArrayList[T](input: List[T]) = new AsArrayList[T](input)
List(1000, 2000, 3000).asArrayList
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