scala - Why does the absence of an else block translate to Unit type return for a function?

Question

I noticed there is a type mismatch caused in the line else if(r1 == 0 || divisors.tail.isEmpty || !divisors.tail.contains(r1)){newAcc}. Because there is no else clause to my if ... else if ...

def euclidianDivision(dividend:Int,divisor:Int):(Int,Int)={
  val quotient = dividend/divisor
  val remainder = dividend%divisor

  (quotient,remainder)
}
def firstExpansion(dividend:Int,divisors:List[Int]):List[(Int,Int)]={
  def firstExpansionIter(dividend:Int,divisors:List[Int], acc:List[(Int,Int)]):List[(Int,Int)]= {
    val div1:Int = divisors.head
    val (q1,r1):(Int,Int) = euclidianDivision(dividend,div1)
    val newAcc:List[(Int,Int)] = acc:::List((div1,q1))
    if (divisors.tail.contains(r1)){
      firstExpansionIter(r1,divisors.tail,newAcc)
    }else if(r1 == 0 || divisors.tail.isEmpty || !divisors.tail.contains(r1)){newAcc}
  }
  firstExpansionIter(dividend,divisors,List((0,0))).tail
}

Here's the error code:

Error:(32, 15) type mismatch; found : Unit required: List[(Int, Int)] }else if(r1 == 0 || divisors.tail.isEmpty || !divisors.tail.contains(r1)){newAcc}

I can correct this by adding the else clause, but how come if there is no outcome handled by default, the function tries to return a Unit?

N.B : Corrected code :

def firstExpansion(dividend:Int,divisors:List[Int]):List[(Int,Int)]={
  def firstExpansionIter(dividend:Int,divisors:List[Int], acc:List[(Int,Int)]):List[(Int,Int)]= {
    val div1:Int = divisors.head
    val (q1,r1):(Int,Int) = euclidianDivision(dividend,div1)
    val newAcc:List[(Int,Int)] = acc:::List((div1,q1))
    if (divisors.tail.contains(r1)){
      firstExpansionIter(r1,divisors.tail,newAcc)
    }else if(r1 == 0 || divisors.tail.isEmpty || !divisors.tail.contains(r1)){newAcc}
    else throw new RuntimeException("Something unexpected happened.")
  }
  firstExpansionIter(dividend,divisors,List((0,0))).tail
}

Answer 1

I can correct this by adding the else clause, but how come if there is no outcome handled by default, the function tries to return a Unit?

In Scala, unlike more "imperative" languages, (almost) everything is an expression (there are very few statements), and every expression evaluates to a value (which also means that every method returns a value).

This means that, for example, the conditional expression if (condition) consequence else differentConsequence is an expression that evaluates to a value.

For example, in this piece of code:

val foo = if (someRandomCondition) 42 else "Hello"

the then part of the expression will evaluate to 42, the else part of the expression will evaluate to "Hello", which means the if expression as a whole will evaluate to either 42 or "Hello".

So, what is the type of foo going to be? Well, in the then case, the value is of type Int and in the else case, the value is of type String. But, this depends on the runtime value of someRandomCondition, which is unknown at compile time. So, the only choice we have as the type for the whole if expression is the lowest common ancestor (technically, the weak least upper bound) of Int and String, which is Any.

In a language with union types, we could give it a more precise type, namely the union type Int | String. (Scala 3 has union types, so we could give the expression this exact type, although Scala 3 will not infer union types.) In Scala 3, we could even annotate it with the even more precise type 42 | "Hello", which is actually the type that TypeScript is going to infer for the equivalent conditional expression:

const foo = someRandomCondition ? 42 : "Hello"

Now, let's move forward towards the code in the question:

val bar = if (someRandomCondition) 42

What is the type of bar going to be? We said above that it is the lowest common ancestor of the types of the then and else branch, but …?what is the type of the else branch? What does the else branch evaluate to?

Remember, we said that every expression evaluates to a value, so the else branch must evaluate to some value. It can't just evaluate to "nothing".

This is solved by a so-called unit value of a unit type. The unit value and type are called the "unit" value and type, because the type is designed in such a way that it can only possibly be inhabited by a single value. The unit type has no members, no properties, no fields, no semantics, no nothing. As such, it is impossible to distinguish two values of the unit type from one another, or put another way: there can only be one value of the unit type, because very other value of the unit type must be identical.

In many programming languages, the unit value and type use the same notation as a tuple value and type, and are simply identified with the empty tuple (). An empty tuple and a unit value are the same thing: they have no content, no meaning. In Haskell, for example, both the type and the value are written ().

Scala also has a unit value, and it is also written (). The unit type, however, is scala.Unit.

So, the unit value, which is a useless value, is used to signify a meaningless return value.

A related, but different concept in some imperative languages is the void type (or in some languages, it is more a "pseudo-type").

Note that "returns nothing" is different from "doesn't return", which will become important in the second part of this answer.

So the first half of the puzzle is: the Scala Language Specification says that

if (condition) expression

is equivalent to

if (condition) expression else ()

Which means that in the (implicit) else case, the return type is Unit, which is not compatible with List[(Int, Int)], and therefore, you get a type error.

But why does throwing an exception fix this?

This brings us to the second special type: Nothing. Nothing is a so-called bottom type, which means that it is a subtype of every type. Nothing does not have any value. So, what then, would a return type of Nothing signify?

It signifies an expression that doesn't return. And I repeat what I said above: this is different from returning nothing.

A method that has only a side-effect returns nothing, but it does return. Its return type is Unit and its return value is (). It doesn't have a meaningful return value.
A method that has an infinite loop or throws an exception doesn't return at all. Its return type is Nothing and it doesn't have a return value.

And that is why throwing an exception in the else clause fixes the problem: this means that the type of the else clause is Nothing, and since Nothing is a subtype of every type, it doesn't even matter what the type of the then clause is, the lowest common supertype of the type of the then clause and Nothing will always be the type of the then clause. (Think about it: the lowest common ancestor of a father and any of his children, grandchildren, great-grandchildren, etc. will always be the father himself. The lowest common ancestor of T and any subtype of T will always be T. Since Nothing is a subtype of all types, the lowest common ancestor of T and Nothing will always be T because Nothing is always a subtype of T, no matter what T is.)