Scala Variance Explained

There are quite a few articles and resources about variance in Scala available on line. Still, I decided to add my own contribute with this post, trying to keep the whole topic as simple as I could, using familiar concept, few diagrams and some code you can play with. That hopefully will make the whole thing a bit easier to grasp. I also decided not to include reference to category theory: not because of not use (on the contrary) but because I hope this rough explanation could work as a sort of encouragement and introduction to it (if you want to dig it further I linked some resources at the end of the page).

Kingdom Animalia

As we need an example, actually an analogy, I chose something that lack originality but not familiarity (hopefully). The code is available here but you won’t find any surprise in it. It’s just a hierarchy of classes modeling the relationship between different groups of animals:
Kingdom Animalia


From the hierarchy above Vertebrate is the parent class of Fish (or Fish a subtype of Vertebrate) and we can clearly write something like:

val vertebrate: Vertebrate = new Fish()

And that’s because a fish is a vertebrate (or at least we modeled it wisely as such). But what about a higher-kinded types like T[A]? Let’s say for example that we have a generic class Cage[A] that takes a type parameter, what can be said about the relationship between Cage[Vertebrate] and Cage[Fish]? Is Cage[Vertebrate] the parent of Cage[Fish]? Does the same relationship between a vertebrate and a fish apply to the respective higher-kinded type?

Variance is mainly about how we answer this question. We have precisely three options: invariance, covariance, and contravariance


In the invariant case the answer to the previous question is pretty simple. There is no relationship between the types and their corresponding higher-kinded types.

An invariant class Cage would be defined in the following way in Scala:

class Cage[A]

Then we can unsurprisingly write val mammalCage: Cage[Mammal] = new Cage[Mammal] but, despite Primate being a subtype of Mammal, the following will not compile:

val primateCage: Cage[Mammal] = new Cage[Primate]
// Expect an error like:
// [error] Note: kingdom.animalia.vertebrate.mammal.primate.Primate <: kingdom.animalia.vertebrate.mammal.Mammal, but class Cage is invariant in type A.
// [error] You may wish to define A as +A instead.

And for analogous reasons val vertebrateCage: Cage[Mammal] = new Cage[Vertebrate] won’t compile either. Conclusion: an invariant class Cage does not preserve the inheritance relationship between its type arguments.


In the covariant case the answer to our initial question is more articulate and interesting. If B is a subtype of A, then even F[B] is a subtype of F[A].

Back to our example, a covariant Cage class in Scala would look like:

class Cage[+A]


val mammalCage: Cage[Mammal] = new Cage[Mammal]
val primateCage: Cage[Mammal] = new Cage[Primate]
val homoCage: Cage[Mammal] = new Cage[Homo]

Will all compile without errors because of the parent/child relationship between mammal/primate (or mammal/homo) is extended to the higher-kinded type Cage so that Cage[Mammal] is the parent of Cage[Primate]. Of course the same won’t be true of:

// expect a type mismatch error if you try to compile any of these
val reptileCage: Cage[Mammal] = new Cage[Reptile]
val vertebrateCage: Cage[Mammal] = new Cage[Vertebrate]

As there is no parent/child relationship between a mammal and a reptile, nor between a mammal and a vertebrate. In this last case though, the opposite is true actually. All mammals are vertebrate and we modeled Vertebrate to be the parent of Mammal and that introduce us to the next topic.


As you might have guessed at this point, contravariance is the opposite of covariance:

We still extend the types’ inheritance to the corresponding higher-kinded type but in this case, so to speak, the arrow is inverted. More precisely, if B is a subtype of A, then F[A] is a subtype of F[B] (note the difference with the covariant case). A contravariant Cage class in Scala would look like:

class Cage[-A]

Then in such a cage for a mammal, we can keep:

val mammalCage: Cage[Mammal] = new Cage[Mammal]
val vertebrateCage: Cage[Mammal] = new Cage[Vertebrate]
val animalCage: Cage[Mammal] = new Cage[Animal]

But the following lines will not compile:

// type mismatch error
val primateCage: Cage[Mammal] = new Cage[Primate]
// type mismatch error
val molluscCage: Cage[Mammal] = new Cage[Mollusc]


  • In an invariant Cage[Mammal] I can keep only a type Mammal (not its subtypes or supertypes)
  • In a covariant Cage[Mammal] I can keep a type Mammal or one of its subtypes (but not its supertypes)
  • In a contravariant Cage[Mammal] I can keep a type Mammal or one of its supertypes (but not its subtypes)

And note that this restriction is imposed at compile time. In other words, the compiler will check for you what can and cannot be passed to an higher-kinded type accordingly to its type arguments variance definition.

Covariant type A occurs in contravariant position

That was a sort of really general overview but It might leave you with some question about the use of variance from a practical perspective, including advantages and disadvantages. In this section (and the following), I’ll try to present some example along with some common Scala class that makes use of variance and I’ll start first with some code that does not compile: a broken cage.

class BrokenCage[+A](var guest: A)

The BrokenCage class is defined as covariant in is type but then we simply pass a guest of type A. The compiler will not digest that and will return an error like covariant type A occurs in contravariant position in type A of value guest.
To explain what is happening here and why the compiler is not happy, an absurd reasoning might help. So let’s assume for a moment that the code above compile and see what happens. I first create a cage with a primate in it:

val primateCage = new BrokenCage[Primate](new Primate)

Because BrokenCage is covariant in A, BrokenCage[Primate] is a subtype of BrokenCage[Animal] and therefore we can assign a BrokenCage[Primate] to a BrokenCage[Animal]

val animalCage: BrokenCage[Animal] = primateCage

So now we have an animalCage of type BrokenCage[Animal]. Then there shouldn’t be any problem to assign an invertebrate as a guest. After all an invertebrate is an animal and BrokenCage is covariant in its type.

animalCage.guest = new Invertebrate

But our animalCage is the primateCage we define at the beginning so basically we managed to put an invertebrate in a cage for a primate! This issue is known as heap pollution and the Scala compiler will prevent these sort of problems from happening, but it can occur for example in a normal Java array:

Primate[] primates = new Primate[5];
Animal[] animals =  primates;
animals[0] = new Invertebrate();

This code will compile and will result in an ArrayStoreException at runtime. One of the advantages of variance in Scala is precisely that these kind of error are captured at compile time rather than at runtime.
And just as a note (as it’s out of scope for the present post), the BrokenCage class can be fixed either using a val (rather than a var) or an upper type bound like in:

class FixedCage[+A, B <: A](var guest: B)

Where we specify that the guest B has to be a subtype of A (there are a couple of simple test around it in the code)


To see how variance (covariance in the specific case) is used in Scala, the familiar Option trait is a good example. A simple implementation would be (and the actual Scala implementation is not much different):

sealed trait Option[+A]

case class Some[+A]( value: A ) extends Option[A]
case object None extends Option[Nothing]

Option is covariant in it’s type A (and so it’s Some). But why is covariant might not be immediately evident so to understand a bit more about it, let’s try to make it invariant and see what sort of behavior we should expect as a result of it:

sealed trait Option[A]

case class Some[A]( value: A ) extends Option[A]
case object None extends Option[Nothing]

The first thing to notice is that even Some has to be invariant in its type argument A as it extends Option and now Option is invariant in A. Besides (not surprisingly) we cannot assign an Option[Primate] to an Option[Mammal].

var optionalMammal: Option[Mammal] = Some(new Mammal)
val optionalPrimate: Option[Primate] = Some(new Primate)
// This won't compile
// optionalMammal = optionalPrimate

Despite Primate being a subtype of Mammal, that relationship doesn’t get passed to our invariant version of Option. Same applies to this invariant version of Some (for the same reason). Perhaps not so obvious, we cannot assign a None to Option[Mammal]:

// This won't compile:
// val mammal: Option[Mammal] = None

Nothing is a subtype of any type in Scala but this time Option is invariant so we can’t assign to an Option[Mammal] anything that is not an Option[Mammal], hence not an Option[Nothing], nor None which extends Option[Nothing]. We could get around this inconvenient with an upper type bound defining Option trait like sealed trait Option[A <: Nothing] but I let you imagine the usefulness of such Option type. More generally, an invariant version of Option fails to meet those expectation an Option type is supposed to deliver (in the code, you can find also a version of a covariant Option with an invariant definition for Some).


Functions are another interesting example of variance usage in the Scala language. The trait Function1 is defined as contravariant in its argument and covariant in its returned type. Simplifying it looks like:

trait Function1[-X,+Y] {
  def apply(arg: X): Y

What we are saying with such type definition is that if A is a supertype of X and B is a subtype of Y, then Function1[A,B] is a subtype of Function1[X,Y]. In terms of the kingdom animalia analogy used so far, Function1[Vertebrate,Homo] is a subtype of Function1[Mammal,Primate] as Vertebrate is a supertype of Mammal and Homo is a subtype of Primate:

Although the reason for such a choice might not be immediately evident. When we think about a class in our animal kingdom it’s quite intuitive what a parent/child relationship involves. Primate is a subtype of Mammal because all Primates are Mammals. Hence, if you ask me for a mammal, I can give you a primate as a primate is a mammal. This concept has a name: Liskov substitution principle. It states (from Wikipedia):

If S is a subtype of T, then objects of type T may be replaced with objects of type S (i.e. an object of type T may be substituted with any object of a subtype S) without altering any of the desirable properties of T

Back to our example, if we want to understand why Function1[Vertebrate,Homo] is a subtype of Function1[Mammal,Primate], or in other words, why Function1 is defined as contravariant in its argument and covariant in its returned type, then we need to start from those desirable properties that the present definition is not supposed to alter (and that, likely, a different definition would have altered instead). For a function (and for a category in category theory) we expect the composition operation as a primitive. Function composition can be defined more rigorously but intuitively it says that if we have a function from A to B and a function from B to C, then we can always compose the two functions and obtain a function from A to C.

Translated in Scala

val f: Vertebrate => Mammal = ???
val g: Mammal => Primate = ???

// We can compose those to obtain a new function Vertebrate => Primate
val fromVertebrateToPrimate: Vertebrate => Primate = f andThen g // or g compose f

Given the definition of Function we saw, can we safely say that replacing a function with one of its subtypes doesn’t alter the compose operation? Let’s start checking whether making the return type covariant is actually a problem.
Assume g stays the same but we replace f with a subtype of f. Accordingly to our definition of function the return type is covariant so Vertebrate => Homo is a subtype of f (as Homo is a subtype of Mammal)


It should be clear from the picture that replacing f with its subtype doesn’t prevent us from compose the two functions. g takes a Mammal as argument and Homo is a Mammal.
Now let’s keep f as it is and replace g with a subtype of g. As a function is contravariant in its argument, a function Vertebrate => Primate is a subtype of g (as the argument Vertebrate is a supertype of Mammal)


Even in this case the composition is preserved. g takes a vertebrate now and all mammals are vertebrate. To complete the picture, remain to be seen how a different definition of function would have impact on the composition operation. What for example if the return type is contravariant? Then we could replace f with a function Vertebrate => Vertebrate (with the return type being a a supertype of Mammal and therefore Vertebrate => Vertebrate a subtype of Vertebrate => Mammal). Would that impact on our ability to compose the two functions?


Yes, as the picture above show. For example we pass a vertebrate to f and f returns a fish (which is a vertebrate but not a mammal). g is not defined for fishes though, only for mammals.
Similar outcome if we define function as being covariant in its argument. In the following example we keep f unchanged but we replace the original g with a subtype: Primate => Primate. If we define Function as covariant in its argument then I can replace Mammal with its subtype Primate and therefore Primate => Primate would be a legitimate subtype of Mammal => Primate accordingly to this definition:


Even in this case our ability to compose the two function is lost as g is not defined for all the possible mammals but just for a specific subset of them: primates. If f returns a dog, g wouldn’t know what to do with it.
The lesson in case of function can be expressed in more formal way introducing the concept of domain and codomain of a function. A domain is the set of values for which the function is defined (its possible arguments) and the codomain is the set of all possible values that the function can return. So, for instance, in a function Mammal => Homo the set of all mammals is the domain of the function and the set of all humans is the codomain. We can now say that given two functions A => B and C => D, those can be composed into a function A => D if and only if the codomain of the first function (B) is a subset of the domain of the second function (C). And that is what is captured in Scala definition of function with its contravariant argument type and covariant returned type.

Code is available here:
And here:

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