CODEWITHSUNDEEP
scalacollections
ScalaBoy
ScalaBoy

Reputation: 3392

How to get Seq[String] instead of Seq[Any]?

I need to do the following transformation and columnNames should be Seq[String], however it is always Seq[Any]. How can I solve this task?

val cols = Seq("col1","col2")
val columnNames: Seq[String] = cols ++ "col3"

Upvotes: 1

Views: 3428

Answers (3)

RAGHHURAAMM
RAGHHURAAMM

Reputation: 1099

++ is the binary operator(method) on two collections. But the :+ or +: is the operator to add an element to the collection( the collection should always be adjacent to : and the element to be added to the collection should always be adjacent to + in :+ ie., it should be used as collection :+ element or element +: collection to append or prepend the element to the collection respectively.So, the cols :+"col3" will give us the expected answer, because then the "col3" will be treated as a string element to be added to cols collection.

While using, cols ++ "col3" , the String "col3" is being treated by the compiler as a collection ie.,as List('c','o','l','3'), cols is a collection of strings. Hence the resulting collection is List[Any], because it contains strings and chars, the super-type of String and Char is inferred to be Any by the compiler.

Upvotes: 1

james.bondu
james.bondu

Reputation: 1162

It should be : val columnNames: Seq[String] = cols :+ "col3"

:+ is a method on whatever type is returned by someVariable.next().

You are getting Class Any is the root of the Scala class hierarchy. Every class in a Scala execution environment inherits directly or indirectly from this class.

Upvotes: 0

Raphael Roth
Raphael Roth

Reputation: 27373

you should do : 

val columnNames: Seq[String] = cols :+ "col3"

or

val columnNames: Seq[String] = cols ++ Seq("col3")

In your case the result is Seq[Any] because Any is the common supertype of String and Char

Upvotes: 2

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