Scala: how to check whether a object is instance of Array

Question

I want to check whether a obj in Scala is of type Array, not caring about whether what type of the values are inside. Is there some easy way to do this?

var x = Array(1,2,3)
x.isInstanceOf[Array[Any]] // which is false

I want it to be true.

Update:

I have a function:

def someFunction(obj: Any) {
    if (obj.isInstanceOf[Int]) {
        // do something
    }
    if (obj.isInstanceOf[String]) {
        // do something
    }
    if (obj.isInstanceOf[Array[Any]) {
        // for each of the obj inside the Array do something
    }
}

prayagupadhyay · Accepted Answer · 2017-06-09 17:20:58Z

7

array Array(1,2,3) is of type Int, so Array[Any] would weirdly return false only in case of Array collection.

If you don't care the type of Array, you can compare to Array[_] (array of whatever type)

scala> var x = Array(1,2,3)
x: Array[Int] = Array(1, 2, 3)

scala> x.isInstanceOf[Array[Int]]
res0: Boolean = true

scala> x.isInstanceOf[Array[_ <: Any]]
res7: Boolean = true

scala> x.isInstanceOf[Array[_ <: AnyVal]]
res12: Boolean = true

scala> x.isInstanceOf[Array[_]]
res13: Boolean = true

scala.Int extends AnyVal which extends Any you can explicitly mentions _ extends AnyVal.

But other scala collections like List[T] is instance of List[Any]

scala> List(1, 2).isInstanceOf[List[Any]]
res30: Boolean = true

scala> Seq(1, 2).isInstanceOf[Seq[Any]]
res33: Boolean = true

Also, List[T] is instance of Seq[Any], because List extends Seq, and T extends Any

scala> List(1, 2).isInstanceOf[Seq[Any]]
res35: Boolean = true

AND, for if else you can solve with pattern match way,

array match {
   case x: Int => println("int") 
   case x: String => "string" 
   case Array[Int] => println("ints")
   case Array[String] => println("strings")
   case _ => println("whatever")
 }

edited Jun 9, 2017 at 17:20

answered Jun 9, 2017 at 9:35

prayagupadhyay

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2 Comments

puhlen Over a year ago

There is a problem with you List example. It will always return true no matter what type you use for T because of erasure. List(1,2,3).isInstanceOf[List[String]] returns true.

prayagupadhyay Over a year ago

You are very right because its matching the erasure, even List(1,2,3).isInstanceOf[List[Map[String, String]]] returns true, but compiler will warn you that List[Int] cannot also be a List[String].

chengpohi · Accepted Answer · 2017-06-09 09:48:18Z

3

x.isInstanceOf[Array[_]]

use _ for container any type match. and for _ it's called existential type.

Document:

Underscore for existential type in Scala

edited Jun 9, 2017 at 9:48

answered Jun 9, 2017 at 9:35

chengpohi

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2 Comments

xidui Over a year ago

Thank you. It works! But I am a little bit confused about the design. It is more intuitive for beginner to use Any here but why we use underscore. I think every beginner will try Array[Any] first instead of Array[_]

Jörg W Mittag Over a year ago

I'm not sure why it is more intuitive to use Any. An Array[X] is an Array[X], it is not an Array[Y]. An Array[Int] is an Array[Int], it is not an Array[Any]. Array[Int] and Array[Any] are two completely different types, they have absolutely nothing to do with each other. _ is used as a wildcard character all over Scala, why would it be confusing to use it as a wildcard character here?

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