Why do I get a sequence of chars when I map a Range to String?

Question
1

I’m just starting to learn scala and I came across the following case in the REPL:

(1 to 3).map(toString)
\\=> IndexedSeq[Char] = Vector(l, i, n)

What I was expecting can be achieved by:

(1 to 3).map(n => n.toString)
\\=>  IndexedSeq[String] = Vector(1, 2, 3)

Strangely, this works as expected

(1 to 3)(0)
\\=>   Int = 1

Just wondering what the reasoning is here?

1 Like
2

Nice puzzler!

What’s happening here is that you’re calling the toString method of the object that is wrapping your code. You can see that by evaluating println(toString). That String is then implicitly converted to a WrappedString because that is a subtype of Int => Char which conforms to the expected type of the argument to the map function. The same happens in this code: "bar"(1) will return 'a' because it’s the character at index 1.

What you probably meant to write is

(1 to 3).map(_.toString)
3 Likes
3

Hi Jasper,

thanks for the quick explanation. I can follow what you wrote in the REPL:

scala> println(toString)
\\=> $line4.$read$$iw@7d3c09ec

and the index values 1 to 3 map to the characters l, i, n, which is more obvious when you increase the length of the range:

(1 to 10).map(toString)
\\=>  IndexedSeq[Char] = Vector(l, i, n, e, 3, 4, ., $, r, e)

You’re right, I should have used the underscore!