How to split strings into characters in Scala

Question

Do you need characters?

"Test".toList    // Makes a list of characters
"Test".toArray   // Makes an array of characters

Do you need bytes?

"Test".getBytes  // Java provides this

Do you need strings?

"Test".map(_.toString)    // Vector of strings
"Test".sliding(1).toList  // List of strings
"Test".sliding(1).toArray // Array of strings

Do you need UTF-32 code points? Okay, that's a tougher one.

def UTF32point(s: String, idx: Int = 0, found: List[Int] = Nil): List[Int] = {
  if (idx >= s.length) found.reverse
  else {
    val point = s.codePointAt(idx)
    UTF32point(s, idx + java.lang.Character.charCount(point), point :: found)
  }
}
UTF32point("Test")

You can use toList as follows:

scala> s.toList         
res1: List[Char] = List(T, e, s, t)

If you want an array, you can use toArray

scala> s.toArray
res2: Array[Char] = Array(T, e, s, t)

Actually you don't need to do anything special. There is already implicit conversion in Predef to WrappedString and WrappedString extends IndexedSeq[Char] so you have all goodies that available in it, like:

"Test" foreach println
"Test" map (_ + "!") 

Edit

Predef has augmentString conversion that has higher priority than wrapString in LowPriorityImplicits. So String end up being StringLike[String], that is also Seq of chars.

Additionally, it should be noted that if what you actually want isn't an actual list object, but simply to do something which each character, then Strings can be used as iterable collections of characters in Scala

for(ch<-"Test") println("_" + ch + "_") //prints each letter on a different line, surrounded by underscores