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) => ??? }
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.
result
mapLet'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) }
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)
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
.
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.
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.