Yield String from List[Char]

Question

I have a l: List[Char] of characters which I want to concat and return as a String in one for loop.

I tried this

val x: String = for(i <- list) yield(i)

leading to

 error: type mismatch;  
 found   : List[Char]  
 required: String

So how can I change the result type of yield?

Thanks!

Is there a reason why you need to use the for comprehension for that? — soc, May 29 '11 at 13:19

score 78 · Answer 1 · answered May 28 '11 at 15:27

78

Try this:

val x: String = list.mkString

This syntax:

for (i <- list) yield i

is syntactic sugar for:

list.map(i => i)

and will thus return an unchanged copy of your original list.

answered May 28 '11 at 15:27

Jean-Philippe Pellet

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@soc: This is indeed possible with scalaz: `(List("a", "b", "c") ∑) assert_≟ "abc"`. – Debilski May 29 '11 at 12:38

score 4 · Answer 2 · edited May 23 '17 at 10:31

4

You can use the following:

val x: String = (for(i <- list) yield(i))(collection.breakOut)

See this question for more information about breakOut.

edited May 23 '17 at 10:31

Community

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answered May 28 '11 at 22:53

Daniel C. Sobral

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score 2 · Answer 3 · answered Mar 11 '15 at 23:46

You can use any of the three mkString overloads. Basically it converts a collection into a flat String by each element's toString method. Overloads add custom separators between each element.

It is a Iterable's method, so you can also use it in Map or Set.

See http://www.scala-lang.org/api/2.7.2/scala/Iterable.html for more details.

Yield String from List[Char]

3 Answers3