Student Questions about Scala, Part 1

Topics: conventions, regexes, mapping, partitioning, vectors vs lists, overloaded constructors, case classes, traits, multiple inheritance, implicits


I’m currently teaching a course on Applied Text Analysis and am using Scala as the programming language taught and used in the course. Rather than creating more tutorials, I figured I’d take a page from Brian Dunning’s playbook on his Skeptoid podcast (highly recommended) when he takes student questions.  So, I had the students in the course submit questions about Scala that they had, based on the readings and assignments thus far. This post covers over half of them — the rest will be covered in a follow up post.

I start with some of the more basic questions, and the questions and/or answers progressively get into more intermediate level topics. Suggestions and comments to improve any of the answers are very welcome!

Basic Questions

Q. Concerning addressing parts of variables: To address individual parts of lists, the numbering of the items is (List 0,1,2 etc.) That is, the first element is called “0”. It seems to be the same for Arrays and Maps, but not for Tuples- to get the first element of a Tuple, I need to use Tuple._1. Why is that?

A. It’s just a matter of convention — tuples have used a 1-based index in other languages like Haskell, and it seems that Scala has adopted the same convention/tradition. See:

Q. It seems that Scala doesn’t recognize the “b” boundary character as a regular expression.  Is there something similar in Scala?

A. Scala does recognize boundary characters. For example, the following REPL session declares a regex that finds “the” with boundaries, and successfully retrieves the three tokens of “the” in the example sentence.

scala> val TheRE = """\bthe\b""".r
TheRE: scala.util.matching.Regex = \bthe\b

scala> val sentence = "She think the man is a stick-in-the-mud, but the man disagrees."
sentence: java.lang.String = She think the man is a stick-in-the-mud, but the man disagrees.

scala> TheRE.findAllIn(sentence).toList
res1: List[String] = List(the, the, the)

Q. Why doesn’t the method “split” work on args? Example: val arg = args.split(” “). Args are strings right, so split should work?

A. The args variable is an Array, so split doesn’t work on them. Arrays are, in effect, already split.

Q. What is the major difference between foo.mapValues(x=>x.length) and>x.length). Some places one works and one does not.

A. The map function works on all sequence types, including Seqs and Maps (note that Maps can be seen as sequences of Tuple2s). The mapValues function, however, only works on Maps. It is essentially a convenience function. As an example, let’s start with a simple Map from Ints to Ints.

scala> val foo = List((1,2),(3,4)).toMap
foo: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 3 -> 4)

Now consider the task of adding 2 to each value in the Map. This can be done with the map function as follows.

scala> { case(key,value) => (key,value+2) }
res5: scala.collection.immutable.Map[Int,Int] = Map(1 -> 4, 3 -> 6)

So, the map function iterates over key/value pairs. We need to match both of them, and then output the key and the changed value to create the new Map. The mapValues function makes this quite a bit easier.

scala> foo.mapValues(2+)
res6: scala.collection.immutable.Map[Int,Int] = Map(1 -> 4, 3 -> 6)

Returning to the question about computing the length using mapValues or map — then it is just a question of which values you are transforming, as in the following examples.

scala> val sentence = "here is a sentence with some words".split(" ").toList
sentence: List[java.lang.String] = List(here, is, a, sentence, with, some, words)

res7: List[Int] = List(4, 2, 1, 8, 4, 4, 5)

scala> val firstCharTokens = sentence.groupBy(x=>x(0))
firstCharTokens: scala.collection.immutable.Map[Char,List[java.lang.String]] = Map(s -> List(sentence, some), a -> List(a), i -> List(is), h -> List(here), w -> List(with, words))

scala> firstCharTokens.mapValues(_.length)
res9: scala.collection.immutable.Map[Char,Int] = Map(s -> 2, a -> 1, i -> 1, h -> 1, w -> 2)

Q. Is there any function that splits a list into two lists with the elements in the alternating positions of the original list? For example,

MainList =(1,2,3,4,5,6)

List1 = (1,3,5)
List2 = (2,4,6)

A. Given the exact main list you provided, one can use the partition function and use the modulo operation to see whether the value is divisible evenly by 2 or not.

scala> val mainList = List(1,2,3,4,5,6)
mainList: List[Int] = List(1, 2, 3, 4, 5, 6)

scala> mainList.partition(_ % 2 == 0)
res0: (List[Int], List[Int]) = (List(2, 4, 6),List(1, 3, 5))

So, partition returns a pair of Lists. The first has all the elements that match the condition and the second has all the ones that do not.

Of course, this wouldn’t work in general for Lists that have Strings, or that don’t have Ints in order, etc. However, the indices of a List are always well-behaved in this way, so we just need to do a bit more work by zipping each element with its index and then partitioning based on indices.

scala> val unordered = List("b","2","a","4","z","8")
unordered: List[java.lang.String] = List(b, 2, a, 4, z, 8)

scala> unordered.zipWithIndex
res1: List[(java.lang.String, Int)] = List((b,0), (2,1), (a,2), (4,3), (z,4), (8,5))

scala> val (evens, odds) = unordered.zipWithIndex.partition(_._2 % 2 == 0)
evens: List[(java.lang.String, Int)] = List((b,0), (a,2), (z,4))
odds: List[(java.lang.String, Int)] = List((2,1), (4,3), (8,5))

res2: List[java.lang.String] = List(b, a, z)

res3: List[java.lang.String] = List(2, 4, 8)

Based on this, you could of course write a function that does this for any arbitrary list.

Q. How to convert a List to a Vector and vice-versa?

A. Use toIndexSeq and toList.

scala> val foo = List(1,2,3,4)
foo: List[Int] = List(1, 2, 3, 4)

scala> val bar = foo.toIndexedSeq
bar: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 2, 3, 4)

scala> val baz = bar.toList
baz: List[Int] = List(1, 2, 3, 4)

scala> foo == baz
res0: Boolean = true

Q. The advantage of a vector over a list is the constant time look-up. What is the advantage of using a list over a vector?

A. A List is slightly faster for operations at the head (front) of the sequence, so if all you are doing is doing a traversal (accessing each element in order, e.g. when mapping), then Lists are perfectly adequate and may be more efficient. They also have some nice pattern matching behavior for case statements.

However, common wisdom seems to be that you should default to using Vectors. See Daniel Spiewak’s nice answer on Stackoverflow:

Q. With splitting strings, holmes.split(“\\s”) – \n and \t just requires a single ‘\’ to recognize its special functionality but why two ‘\’s are required for white space character?

A. That’s because \n and \t actually mean something in a String.

scala> println("Here is a line with a tab\tor\ttwo, followed by\na new line.")
Here is a line with a tab    or    two, followed by
a new line.

scala> println("This will break\s.")
<console>:1: error: invalid escape character
println("This will break\s.")

So, you are supplying a String argument to split, and it uses that to construct a regular expression. Given that \s is not a string character, but is a regex metacharacter, you need to escape it. You can of course use split(“””\s”””), though that isn’t exactly better in this case.

Q. I have long been programming in C++ and Java. Therefore, I put semicolon at the end of the line unconsciously. It seems that the standard coding style of Scala doesn’t recommend to use semicolons. However, I saw that there are some cases that require semicolons as you showed last class. Is there any specific reason why semicolon loses its role in Scala?

A. The main reason is to improve readability since the semicolon is rarely needed when writing standard code in editors (as opposed to one liners in the REPL). However, when you want to do something in a single line, like handling multiple cases, you need the semicolons.

scala> val foo = List("a",1,"b",2)
foo: List[Any] = List(a, 1, b, 2)

scala> { case(x: String) => x; case(x: Int) => x.toString }
res5: List[String] = List(a, 1, b, 2)

But, in general, it’s best to just split these cases over multiple lines in any actual code.

Q. Is there no way to use _ in map like methods for collections that consist of pairs? For example, List((1,1),(2,2)).map(e => e._1 + e._2) works, but List((1,1),(2,2)).map(_._1 + _._2) does not work.

A. The scope in which the _ remains unanambigious runs out past its first invocation, so you only get to use it once. It is better anyway to use a case statement that makes it clear what the members of the pairs are.

scala>  List((1,1),(2,2)).map { case(num1, num2) => num1+num2 }
res6: List[Int] = List(2, 4)

Q. I am unsure about the exact meaning of and the difference between “=>” and “->”. They both seem to mean something like “apply X to Y” and I see that each is used in a particular context, but what is the logic behind that?

A. The use of -> simply constructs a Tuple2, as is pretty clear in the following snippet.

scala> val foo = (1,2)
foo: (Int, Int) = (1,2)

scala> val bar = 1->2
bar: (Int, Int) = (1,2)

scala> foo == bar
res11: Boolean = true

Primarily, it is syntactic sugar that provides an intuitive symbol for creating elements of a a Map. Compare the following two ways of declaring the same Map.

scala> Map(("a",1),("b",2))
res9: scala.collection.immutable.Map[java.lang.String,Int] = Map(a -> 1, b -> 2)

scala> Map("a"->1,"b"->2)
res10: scala.collection.immutable.Map[java.lang.String,Int] = Map(a -> 1, b -> 2)

The second seems more readable to me.

The use of => indicates that you are defining a function. The basic form is ARGUMENTS => RESULT.

scala> val addOne = (x: Int) => x+1
addOne: Int => Int = <function1>

scala> addOne(2)
res7: Int = 3

scala> val addTwoNumbers = (num1: Int, num2: Int) => num1+num2
addTwoNumbers: (Int, Int) => Int = <function2>

scala> addTwoNumbers(3,5)
res8: Int = 8

Normally, you use it in defining anonymous functions as arguments to functions like map, filter, and such.

Q. Is there a more convenient way of expressing vowels as [AEIOUaeiou] and consonants as [BCDFGHJKLMNPQRSTVWXYZbcdfghjklmnpqrstvwxyz] in RegExes?

A. You can use Strings when defining regexes, so you can have a variable for vowels and one for consonants.

scala> val vowel = "[AEIOUaeiou]"
vowel: java.lang.String = [AEIOUaeiou]

scala> val consonant = "[BCDFGHJKLMNPQRSTVWXYZbcdfghjklmnpqrstvwxyz]"
consonant: java.lang.String = [BCDFGHJKLMNPQRSTVWXYZbcdfghjklmnpqrstvwxyz]

scala> val MyRE = ("("+vowel+")("+consonant+")("+vowel+")").r
MyRE: scala.util.matching.Regex = ([AEIOUaeiou])([BCDFGHJKLMNPQRSTVWXYZbcdfghjklmnpqrstvwxyz])([AEIOUaeiou])

scala> val MyRE(x,y,z) = "aJE"
x: String = a
y: String = J
z: String = E

Q. The “\b” in RegExes marks a boundary, right? So, it also captures the “-“. But if I have a single string “sdnfeorgn”, it does NOT capture the boundaries of that, is that correct? And if so, why doesn’t it?

A. Because there are no boundaries in that string!

Intermediate questions

Q. The flatMap function takes lists of lists and merges them to single list. But in the example:

scala> (1 to 10)>squareOddNumber(x))
res16: List[Option[Int]] = List(Some(1), None, Some(9), None, Some(25), None, Some(49), None, Some(81), None)

scala> (1 to 10).toList.flatMap(x=>squareOddNumber(x))
res17: List[Int] = List(1, 9, 25, 49, 81)

Here it is not list of list but just a list. In this case it expects the list to be Option list.
I tried running the code with function returning just number or None. It showed error. So is there any way to use flatmap without Option lists and just list. For example, List(1, None, 9, None, 25) should be returned as List(1, 9, 25).

A. No, this won’t work because List(1, None, 9, None, 25) mixes Options with Ints.

scala> val mixedup = List(1, None, 9, None, 25)
mixedup: List[Any] = List(1, None, 9, None, 25)

So, you should have your function return an Option which means returning Somes or Nones. Then flatMap will work happily.

One way of think of Options is that they are like Lists with zero or one element, as can be noted by the parallels in the following snippet.

scala> val foo = List(List(1),Nil,List(3),List(6),Nil)
foo: List[List[Int]] = List(List(1), List(), List(3), List(6), List())

scala> foo.flatten
res12: List[Int] = List(1, 3, 6)

scala> val bar = List(Option(1),None,Option(3),Option(6),None)
bar: List[Option[Int]] = List(Some(1), None, Some(3), Some(6), None)

scala> bar.flatten
res13: List[Int] = List(1, 3, 6)

Q. Does scala have generic templates (like C++, Java)? eg. in C++, we can use vector<int>, vector<string> etc. Is that possible in scala? If so, how?

A. Yes, every collection type is parameterized. Notice that each of the following variables is parameterized by the type of the elements they are initialized with.

scala> val foo = List(1,2,3)
foo: List[Int] = List(1, 2, 3)

scala> val bar = List("a","b","c")
bar: List[java.lang.String] = List(a, b, c)

scala> val baz = List(true, false, true)
baz: List[Boolean] = List(true, false, true)

You can create your own parameterized classes straightforwardly.

scala> class Flexible[T] (val data: T)
defined class Flexible

scala> val foo = new Flexible(1)
foo: Flexible[Int] = Flexible@7cd0570e

scala> val bar = new Flexible("a")
bar: Flexible[java.lang.String] = Flexible@31b6956f

scala> val baz = new Flexible(true)
baz: Flexible[Boolean] = Flexible@5b58539f

res0: Int = 1

res1: java.lang.String = a

res2: Boolean = true

Q. How can we easily create, initialize and work with multi-dimensional arrays (and dictionaries)?

A. Use the fill function of the Array object to create them.

scala> Array.fill(2)(1.0)
res8: Array[Double] = Array(1.0, 1.0)

scala> Array.fill(2,3)(1.0)
res9: Array[Array[Double]] = Array(Array(1.0, 1.0, 1.0), Array(1.0, 1.0, 1.0))

scala> Array.fill(2,3,2)(1.0)
res10: Array[Array[Array[Double]]] = Array(Array(Array(1.0, 1.0), Array(1.0, 1.0), Array(1.0, 1.0)), Array(Array(1.0, 1.0), Array(1.0, 1.0), Array(1.0, 1.0)))

Once you have these in hand, you can iterate over them as usual.

scala> val my2d = Array.fill(2,3)(1.0)
my2d: Array[Array[Double]] = Array(Array(1.0, 1.0, 1.0), Array(1.0, 1.0, 1.0))

scala> =>>x+1))
res11: Array[Array[Double]] = Array(Array(2.0, 2.0, 2.0), Array(2.0, 2.0, 2.0))

For dictionaries (Maps), you can use mutable HashMaps to create an empty Map and then add elements to it. For that, see this blog post:

Q. Is the apply function similar to constructor in C++, Java? Where will the apply function be practically used? Is it for intialising values of attributes?

A. No, the apply function is like any other function except that it allows you to call it without writing out “apply”. Consider the following class.

class AddX (x: Int) {
  def apply(y: Int) = x+y
  override def toString = "My number is " + x

Here’s how we can use it.

scala> val add1 = new AddX(1)
add1: AddX = My number is 1

scala> add1(4)
res0: Int = 5

scala> add1.apply(4)
res1: Int = 5

scala> add1.toString
res2: java.lang.String = My number is 1

So, the apply method is just (very handy) syntactic sugar that allows you to specify one function as fundamental to a class you have designed (actually, you can have multiple apply methods as long as each one has a unique parameter list). For example, with Lists, the apply method returns the value at the index provided, and for Maps it returns the value associated with the given key.

scala> val foo = List(1,2,3)
foo: List[Int] = List(1, 2, 3)

scala> foo(2)
res3: Int = 3

scala> foo.apply(2)
res4: Int = 3

scala> val bar = Map(1->2,3->4)
bar: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 3 -> 4)

scala> bar(1)
res5: Int = 2

scala> bar.apply(1)
res6: Int = 2

Q. In the SBT tutorial you discuss “Node” and “Value” as being case classes. What is the alternative to a case class?

A. A normal class. Case classes are the special case. They do two things (and more) for you. The first is that you don’t have to use “new” to create a new object. Consider the following otherwise identical classes.

scala> class NotACaseClass (val data: Int)
defined class NotACaseClass

scala> case class IsACaseClass (val data: Int)
defined class IsACaseClass

scala> val foo = new NotACaseClass(4)
foo: NotACaseClass = NotACaseClass@a5c0f8f

scala> val bar = IsACaseClass(4)
bar: IsACaseClass = IsACaseClass(4)

That may seem like a little thing, but it can significantly improve code readability. Consider creating Lists within Lists within Lists if you had to use “new” all the time, for example. This is definitely true for Node and Value, which are used to build trees.

Case classes also support matching, as in the following.

scala> val IsACaseClass(x) = bar
x: Int = 4

A normal class cannot do this.

scala> val NotACaseClass(x) = foo
<console>:13: error: not found: value NotACaseClass
val NotACaseClass(x) = foo
<console>:13: error: recursive value x needs type
val NotACaseClass(x) = foo

If you mix the case class into a List and map over it, you can match it like you can with other classes, like Lists and Ints. Consider the following heterogeneous List.

scala> val stuff = List(IsACaseClass(3), List(2,3), IsACaseClass(5), 4)
stuff: List[Any] = List(IsACaseClass(3), List(2, 3), IsACaseClass(5), 4)

We can convert this to a List of Ints by processing each element according to its type by matching.

scala> { case List(x,y) => x; case IsACaseClass(x) => x; case x: Int => x }
<console>:13: warning: match is not exhaustive!
missing combination              *           Nil             *             * { case List(x,y) => x; case IsACaseClass(x) => x; case x: Int => x }

warning: there were 1 unchecked warnings; re-run with -unchecked for details
res10: List[Any] = List(3, 2, 5, 4)

If you don’t want to see the warning in the REPL, add a case for things that don’t match that throws a MatchError.

scala> { case List(x,y) => x; case IsACaseClass(x) => x; case x: Int => x; case _ => throw new MatchError }
warning: there were 1 unchecked warnings; re-run with -unchecked for details
res13: List[Any] = List(3, 2, 5, 4)

Better yet, return Options (using None for the unmatched case) and flatMapping instead.

scala> stuff.flatMap { case List(x,y) => Some(x); case IsACaseClass(x) => Some(x); case x: Int => Some(x); case _ => None }
warning: there were 1 unchecked warnings; re-run with -unchecked for details
res14: List[Any] = List(3, 2, 5, 4)

Q. In C++ the default access specifier is private; in Java one needs to specify private or public for each class member where as in Scala the default access specifier for a class is public. What could be the design motivation behind this when one of the purpose of the class is data hiding?

A. The reason is that Scala has a much more refined access specification scheme than Java that makes public the rational choice. See the discussion here:

Another key aspecte of this is that the general emphasis in Scala is on using immutable data structures, so there isn’t any danger of someone changing the internal state of your objects if you have designed them in this way. This in turn gets rid of the ridiculous getter and setter methods that breed and multiply in Java programs. See “Why getters and setters are evil” for more discussion:

After you get used to programming in Scala, the whole getter/setter thing that is so common in Java code is pretty much gag worthy.

In general, it is still a good idea to use private[this] as a modifier to methods and variables whenever they are only needed by an object itself.

Q. How do we define overloaded constructors in Scala?

Q. The way a class is defined in Scala introduced in the tutorial, seems to have only one constructor. Is there any way to provide multiple constructors like Java?

A. You can add additional constructors with this declarations.

class SimpleTriple (x: Int, y: Int, z: String) {
  def this (x: Int, z: String) = this(x,0,z)
  def this (x: Int, y: Int) = this(x,y,"a")
  override def toString = x + ":" + y + ":" + z

scala> val foo = new SimpleTriple(1,2,"hello")
foo: SimpleTriple = 1:2:hello

scala> val bar = new SimpleTriple(1,"goodbye")
bar: SimpleTriple = 1:0:goodbye

scala> val baz = new SimpleTriple(1,3)
baz: SimpleTriple = 1:3:a

Notice that you must supply an initial value for every one of the parameters of the class. This contrasts with Java, which allows you to leave some fields uninitialized (and which tends to lead to nasty bugs and bad design).

Note that you can also provide defaults to parameters.

class SimpleTripleWithDefaults (x: Int, y: Int = 0, z: String = "a") {
  override def toString = x + ":" + y + ":" + z

scala> val foo = new SimpleTripleWithDefaults(1)
foo: SimpleTripleWithDefaults = 1:0:a

scala> val bar = new SimpleTripleWithDefaults(1,2)
bar: SimpleTripleWithDefaults = 1:2:a

However, you can’t omit a middle parameter while specifying the last one.

scala> val foo = new SimpleTripleWithDefaults(1,"xyz")
<console>:12: error: type mismatch;
found   : java.lang.String("xyz")
required: Int
Error occurred in an application involving default arguments.
val foo = new SimpleTripleWithDefaults(1,"xyz")

But, you can name the parameters in the initialization if you want to be able to do this.

scala> val foo = new SimpleTripleWithDefaults(1,z="xyz")
foo: SimpleTripleWithDefaults = 1:0:xyz

You then have complete freedom to change the parameters around.

scala> val foo = new SimpleTripleWithDefaults(z="xyz",x=42,y=3)
foo: SimpleTripleWithDefaults = 42:3:xyz

Q. I’m still not clear on the difference between classes and traits.  I guess I see a conceptual difference but I don’t really understand what the functional difference is — how is creating a “trait” different from creating a class with maybe fewer methods associated with it?

A. Yes, they are different. First off, traits are abstract, which means you cannot create any members. Consider the following contrast.

scala> class FooClass
defined class FooClass

scala> trait FooTrait
defined trait FooTrait

scala> val fclass = new FooClass
fclass: FooClass = FooClass@1b499616

scala> val ftrait = new FooTrait
<console>:8: error: trait FooTrait is abstract; cannot be instantiated
val ftrait = new FooTrait

You can extend a trait to make a concrete class, however.

scala> class FooTraitExtender extends FooTrait
defined class FooTraitExtender

scala> val ftraitExtender = new FooTraitExtender
ftraitExtender: FooTraitExtender = FooTraitExtender@53d26552

This gets more interesting if the trait has some methods, of course. Here’s a trait, Animal, that declares two abstract methods, makeNoise and doBehavior.

trait Animal {
  def makeNoise: String
  def doBehavior (other: Animal): String

We can extend this trait with new class definitions; each extending class must implement both of these methods (or else be declared abstract).

case class Bear (name: String, defaultBehavior: String = "Regard warily...") extends Animal {
  def makeNoise = "ROAR!"
  def doBehavior (other: Animal) = other match {
    case b: Bear => makeNoise + " I'm " + name + "."
    case m: Mouse => "Eat it!"
    case _ => defaultBehavior
  override def toString = name

case class Mouse (name: String) extends Animal {
  def makeNoise = "Squeak?"
  def doBehavior (other: Animal) = other match {
    case b: Bear => "Run!!!"
    case m: Mouse => makeNoise + " I'm " + name + "."
    case _ => "Hide!"
  override def toString = name

Notice that Bear and Mouse have different parameter lists, but both can be Animals because they fully implement the Animal trait. We can now start creating objects of the Bear and Mouse classes and have them interact. We don’t need to use “new” because they are case classes (and this also allowed them to be used in the match statements of the doBehavior methods).

val yogi = Bear("Yogi", "Hello!")
val baloo = Bear("Baloo", "Yawn...")
val grizzly = Bear("Grizzly")
val stuart = Mouse("Stuart")

println(yogi + ": " + yogi.makeNoise)
println(stuart + ": " + stuart.makeNoise)
println("Grizzly to Stuart: " + grizzly.doBehavior(stuart))

We can also create a singleton object that is of the Animal type by using the following declaration.

object John extends Animal {
  def makeNoise = "Hullo!"
  def doBehavior (other: Animal) = other match {
    case b: Bear => "Nice bear... nice bear..."
    case _ => makeNoise
  override def toString = "John"

Here, John is an object, not a class. Because this object implements the Animal trait, it successfully extends it and can act as an Animal. This means that a Bear like baloo can interact with John.

println("Baloo to John: " + baloo.doBehavior(John))

The output of the above code when run as a script is the following.

Yogi: ROAR!
Stuart: Squeak?
Grizzly to Stuart: Eat it!
Baloo to John: Yawn…

The closer distinction is between traits and abstract classes. In fact, everything shown above could have been done with Animal as an abstract class rather than as a trait. One difference is that an abstract class can have a constructor while traits cannot. Another key difference between them is that traits can be used to support limited multiple inheritance, as shown in the next question/answer.

Q. Does Scala support multiple inheritance?

A. Yes, via traits with implementations of some methods. Here’s an example, with a trait Clickable that has an abstract (unimplemented) method getMessage, an implemented method click, and a private, reassignable variable numTimesClicked (the latter two show clearly that traits are different from Java interfaces).

trait Clickable {
  private var numTimesClicked = 0
  def getMessage: String
  def click = {
    val output = numTimesClicked + ": " + getMessage
    numTimesClicked += 1

Now let’s say we have a MessageBearer class (that we may have wanted for entirely different reasons having nothing to do with clicking).

class MessageBearer (val message: String) {
  override def toString = message

A new class can be now created by extending MessageBearer and “mixing in” the Clickable trait.

class ClickableMessageBearer(message: String) extends MessageBearer(message) with Clickable {
  def getMessage = message

ClickableMessageBearer now has the abilities of both MessageBearers (which is to be able to retrieve its message) and Clickables.

scala> val cmb1 = new ClickableMessageBearer("I'm number one!")
cmb1: ClickableMessageBearer = I'm number one!

scala> val cmb2 = new ClickableMessageBearer("I'm number two!")
cmb2: ClickableMessageBearer = I'm number two!

res3: java.lang.String = 0: I'm number one!

scala> cmb1.message
res4: String = I'm number one!

res5: java.lang.String = 1: I'm number one!

res6: java.lang.String = 0: I'm number two!

res7: java.lang.String = 2: I'm number one!

res8: java.lang.String = 1: I'm number two!

Q. Why are there toString, toInt, and toList functions, but there isn’t a toTuple function?

A. This is a basic question that leads directly to the more advanced topic of implicits. There are a number of reasons behind this. To start with, it is important to realize that there are many types of Tuples, starting with a Tuple with a single element (a Tuple1) up to 22 elements (a Tuple22). Note that when you use (,) to create a tuple, it is implicitly invoking a constructor for the corresponding TupleN of the correct arity.

scala> val b = (1,2,3)
b: (Int, Int, Int) = (1,2,3)

scala> val c = Tuple3(1,2,3)
c: (Int, Int, Int) = (1,2,3)

scala> b==c
res4: Boolean = true

Given this, it is obviously not meaningful to have a function toTuple on Seqs (sequences) that are longer than 22. This means there is no generic way to have, say a List or Array, and then call toTuple on it and expect reliable behavior to happen.

However, if you want this functionality (even though limited by the above constraint of 22 elements max), Scala allows you to “add” methods to existing classes by using implicit definitions. You can find lots of discussions about implicits by search for “scala implicits”. But, here’s an example that shows how it works for this particular case.

val foo = List(1,2)
val bar = List(3,4,5)
val baz = List(6,7,8,9)


class TupleAble[X] (elements: Seq[X]) {
  def toTuple = elements match {
    case Seq(a) => Tuple1(a)
    case Seq(a,b) => (a,b)
    case Seq(a,b,c) => (a,b,c)
    case _ => throw new RuntimeException("Sequence too long to be handled by toTuple: " + elements)


implicit def seqToTuple[X](x: Seq[X]) = new TupleAble(x)


If you put this into the Scala REPL, you’ll see that the first invocation of foo.toTuple gets an error:

scala> foo.toTuple
<console>:9: error: value toTuple is not a member of List[Int]

Note that class TupleAble takes a Seq in its constructor and then provides the method toTuple, using that Seq. It is able to do so for Seqs with 1, 2 or 3 elements, and above that it throws an exception. (We could of course keeping listing more cases out and go up to 22 element tuples, but this shows the point.)

The second invocation of foo.toTuple still doesn’t work — and that is because foo is a List (a kind of Seq) and there isn’t a toTuple method for Lists. That’s where the implicit function seqToTuple comes in — once it is declared, Scala notes that you are trying to call toTuple on a Seq, notes that there is no such function for Seqs, but sees that there is an implicit conversion from Seqs to TupleAbles via seqToTuple, and then it sees that TupleAble has a toTuple method. Based on that, it compiles and the produces the desired behavior. This is a very handy ability of Scala that can really simplify your code if you use it well and with care.

Copyright 2012 Jason Baldridge

The text of this post is licensed under the Creative Commons Attribution-NonCommercial-ShareAlike License. Attribution may be provided by linking to and to this original post.

Suggestions, improvements, extensions and bug fixes welcome — please email Jason at or provide a comment to this post.

  1. Here are a few comments. The questions are repeated, followed by my comment on your answer.

    Q. Is there any function that splits a list into two lists with the elements in the alternating positions of the original list?

    An alternative that I find more intuitive:

    unordered.grouped(2).map{case Seq(a,b) => (a,b)}.toList.unzip

    Q. How to convert a List to a Vector and vice-versa?

    toIndexedSeq doesn’t necessarily produce a Vector:

    scala> val a = (1 to 5)
    a: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5)

    scala> a.toIndexedSeq
    res9: scala.collection.immutable.IndexedSeq[Int] = Range(1, 2, 3, 4, 5)

    This may be good enough, but different IndexedSeq implementations have different properties. To be sure you’re getting a Vector, try one of these two ways:

    scala> Vector(a: _*)
    res20: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

    scala> Vector() ++ a
    res21: scala.collection.immutable.Vector[Int] = Vector(1, 2, 3, 4, 5)

    Q. Why are there toString, toInt, and toList functions, but there isn’t a toTuple function?

    This is a nice idea. Unfortunately, while your implicit definition compiles, the return type of `toTuple` is not what you want; it’s `Product with Serializable`, which is pretty much uselessly vague. The problem is that each `TupleN` is its own class, so you can’t have a single method that adapts based on the situation because the return type can never be right. Here’s an easier way:

    class TupleAble(elements: Seq[T]) {
    def toTuple2 = elements match { case Seq(a, b) => (a, b) }
    def toTuple3 = elements match { case Seq(a, b, c) => (a, b, c) }
    implicit def seqToTuple(elements: Seq[T]) = new TupleAble(elements)

    val x = List(1, 2).toTuple2

    I see nothing wrong with having `toTupleN` since, in order to have the types work out, you need to know `N` anyway, so there’s really no harm in having it be part of the method name.

    By the way, this makes my answer to the first question above even simpler:


  2. Hey there! I’ve been reading your blog for a while now and finally got the bravery to go ahead and give you a shout out from
    Houston Texas! Just wanted to tell you keep up the good work!

    • Thanks! Haven’t had much time for blogging lately, though that may change in the near future as I do some blogging for my startup, People Pattern.

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