(Scala) How to convert List into a Seq

Question

I have a function like

def myFunction(i:Int*) = i.map(a => println(a))

but I have a List of Int's.

val myList:List[Int] = List(1,2,3,4)

Desired output:
1
2
3
4

How can I programmatically convert myList so it can be inserted into myFunction?

Unrelated: If you are using a side effecting function like println you can use foreach instead of map. — Peter Neyens
– Peter Neyens, Commented Jul 11, 2016 at 22:48
Your question is not really "how to convert a List to a Seq", because the answer to that would be: call toSeq on the list. It's about how to invoke a varargs method when you have a List. — Jesper
– Jesper, Commented Jul 12, 2016 at 11:58

insan-e · Accepted Answer · 2021-04-08 18:05:39Z

4

Looking at your desired input and output, you want to pass a List where a varargs argument is expected.
A varargs method can receive zero or more arguments of same type.
The varargs parameter has type of Array[T] actually.
But you can pass any Seq to it, by using "varargs ascription/expansion" (IDK if there is an official name for this):

myFunction(myList: _*)

edited Apr 8, 2021 at 18:05

answered Jul 11, 2016 at 22:39

insan-e

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3 Comments

Philip Nguyen Over a year ago

Thank you. It would have taken me eons to figure this out on my own.

Manu Chadha Over a year ago

could you please explain your answer as well?

insan-e Over a year ago

@ManuChadha I elaborated bit more on it. I actually expected it to be closed as duplicate... LOL

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