val and var

scala> val a = 123
a: Int = 123

scala> a = 456
<console>:8: error: reassignment to val
       a = 456

scala> var b = 123
b: Int = 123

scala> b = 321
b: Int = 321

• val references are unchangeable: like a final variable in Java, once it has been initialized you cannot change it
• var references are reassignable as a simple variable declaration in Java

Immutable and Mutable collections

  val mut = scala.collection.mutable.Map.empty[String, Int]
  mut += ("123" -> 123)
  mut += ("456" -> 456)
  mut += ("789" -> 789)

  val imm = scala.collection.immutable.Map.empty[String, Int]
  imm + ("123" -> 123)
  imm + ("456" -> 456)
  imm + ("789" -> 789)

  scala> mut
    Map(123 -> 123, 456 -> 456, 789 -> 789)

  scala> imm
    Map()

scala> imm + ("123" -> 123) + ("456" -> 456) + ("789" -> 789)
    Map(123 -> 123, 456 -> 456, 789 -> 789)

The Scala standard library offers both immutable and mutable data structures, not the reference to it. Each time an immutable data structure get "modified", a new instance is produced instead of modifying the original collection in-place. Each instance of the collection may share significant structure with another instance.

Mutable and Immutable Collection (Official Scala Documentation)

Let's pick as an example a function that takes 2 Map and return a Map containing every element in ma and mb:

def merge2Maps(ma: Map[String, Int], mb: Map[String, Int]): Map[String, Int]

A first attempt could be iterating through the elements of one of the maps using for ((k, v) <- map) and somehow return the merged map.

def merge2Maps(ma: ..., mb: ...): Map[String, Int] = {

  for ((k, v) <- mb) {
    ???
  }

}

This very first move immediately add a constrain: a mutation outside that for is now needed. This is more clear when de-sugaring the for:

// this:
for ((k, v) <- map) { ??? }

// is equivalent to:
map.foreach { case (k, v) => ??? }

"Why we have to mutate?"

foreach relies on side-effects. Every time we want something to happen within a foreach we need to "side-effect something", in this case we could mutate a variable var result or we can use a mutable data structure.

Creating and filling the result map

Let's assume the ma and mb are scala.collection.immutable.Map, we could create the result Map from ma:

val result = mutable.Map() ++ ma

Then iterate through mb adding its elements and if the key of the current element on ma already exist, let's override it with the mb one.

mb.foreach { case (k, v) => result += (k -> v) }

Mutable implementation

So far so good, we "had to use mutable collections" and a correct implementation could be:

def merge2Maps(ma: Map[String, Int], mb: Map[String, Int]): Map[String, Int] = {
  val result = scala.collection.mutable.Map() ++ ma
  mb.foreach { case (k, v) => result += (k -> v) }
  result.toMap // to get back an immutable Map
}

As expected:

scala> merge2Maps(Map("a" -> 11, "b" -> 12), Map("b" -> 22, "c" -> 23))
  Map(a -> 11, b -> 22, c -> 23)

Folding to the rescue

How can we get rid of foreach in this scenario? If all we what to do is basically iterate over the collection elements and apply a function while accumulating the result on option could be using .foldLeft:

def merge2Maps(ma: Map[String, Int], mb: Map[String, Int]): Map[String, Int] = {
  mb.foldLeft(ma) { case (result, (k, v)) => result + (k -> v) }
  // or more concisely mb.foldLeft(ma) { _ + _ }
}

In this case our "result" is the accumulated value starting from ma, the zero of the .foldLeft.

Intermediate result

Obviously this immutable solution is producing and destroying many Map instances while folding, but it is worth mentioning that those instances are not a full clone of the Map accumulated but instead are sharing significant structure (data) with the existing instance.

Easier reasonability

It is easier to reason about the semantic if it is more declarative as the .foldLeft approach. Using immutable data structures could help making our implementation easier to reason on.