Can't call overloaded Java method from Scala generic method

Question

I am trying to wrap java.util.Array.binarySearch in a generic fashion in Scala, but the following code doesn't work:

 def binarySlice[T](minValue: T, array: Array[T]): Array[T] = {
   val i = java.util.Arrays.binarySearch(array, minValue)
   val idx = if (i > 0) i else -i - 1
   array.slice(idx, array.length)
 }

The error is: Cannot resolve overloaded method 'binarySearch'. What should I do to make this code work?

Which binarySearch method are you trying to use? There are two implementations with Array[T] but they need another parameters. — Sebastian Celestino
– Sebastian Celestino, Commented Sep 26, 2018 at 17:54
The one with only 2 arguments, the second one being the target value? — Clovis
– Clovis, Commented Sep 26, 2018 at 19:22
There are two binarySearch methods with Array[T] argument, none of them have 2 arguments. public static <T> int binarySearch(T[] a, T key, Comparator<? super T> c) and public static <T> int binarySearch(T[] a, int fromIndex, int toIndex, T key, Comparator<? super T> c) — Sebastian Celestino
– Sebastian Celestino, Commented Sep 26, 2018 at 19:26

Andrzej Jozwik · Accepted Answer · 2018-09-26 18:02:08Z

1

See: Scala replacement for Arrays.binarySearch:

def binarySlice[T <: AnyRef](minValue: T, array: Array[T]): Array[T] = {
   val i = java.util.Arrays.binarySearch(array.asInstanceOf[Array[AnyRef]], minValue)
   val idx = if (i > 0) i else -i - 1
   array.slice(idx, array.length)
 }

You can not use it for primitives. It is work-around for int:

scala> binarySlice[java.lang.Integer](3, Array(0,3,7,8))
res6: Array[Integer] = Array(3, 7, 8)

edited Sep 26, 2018 at 18:02

answered Sep 26, 2018 at 17:54

Andrzej Jozwik

14.8k3 gold badges61 silver badges70 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Can't call overloaded Java method from Scala generic method

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related