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如何将Java流转换为Scala数组?

scala-collections implicit-conversion generics scala java

1

Abhijit Sarkar · 技术社区 · 6 年前

鉴于

new Scanner(is)
    .tokens()
    .map(_.toInt)
    .toArray[Int]((_: Int) => Array.ofDim[Int](100000))

Error:(40, 12) no type parameters for method map: (x$1: java.util.function.Function[_ >: String, _ <: R])java.util.stream.Stream[R] exist so that it can be applied to arguments (java.util.function.Function[String,Int])
 --- because ---
argument expression's type is not compatible with formal parameter type;
 found   : java.util.function.Function[String,Int]
 required: java.util.function.Function[_ >: String, _ <: ?R]
Note: String <: Any, but Java-defined trait Function is invariant in type T.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
          .map(_.toInt)
Error:(40, 18) type mismatch;
 found   : java.util.function.Function[String,Int]
 required: java.util.function.Function[_ >: String, _ <: R]
          .map(_.toInt)

1 回复 | 直到 6 年前

1

Abhijit Sarkar 6 年前

编译器正在尝试查找 R 的类型参数 Stream#map map 是一个 Function[String,Int] ,并且它必须与 Function[_ >: String, _ <: R] 地图 ),这是有限制的 Int <: R R <: Object (至少我认为是这样的)。所以没有合适的方法 R

要修复,正如我在评论中提到的,您可以使用 mapToInt ,但实际上还有一个更糟糕的错误消息,它肯定应该被修复(我报告了一个问题):指定 map[Int] 导致

type mismatch;
 found   : Array[Int]
 required: Array[Int]
Note: Int >: Int, but class Array is invariant in type T.
You may wish to investigate a wildcard type such as `_ >: Int`. (SLS 3.2.10)