Type variance explained

Reasoning about subtype relations between parameterised types tend to confuse developers. If you are a Java developer you've probably seen the syntax ? extends T and ? super T, or might even used it to create flexible method signatures. I was very confused at first when I tried to use them as their meaning wasn't clear to me. So, I just placed extends everywhere, and changed it to super when the compiler complained. Where is it even valid to place only super types of some T? It made no sense to me at first. But as it turns out, it makes all the sense. Let's start at the beginning. Say you have a Cat class extending the Animal class. Here, Cat is the subtype of Animal, by definition.

Now, let's create four XInYOut classes, each of them having a single method apply, and use these two classes with them. It doesn't matter what this apply method does, we only care about its signature: it takes one input parameter, and returns an output. Why do we create four? To cover all combinations of the Animal and Cat classes, in the input and output parameters. Without generic parameters yet, what are possible subtyping relations between these classes? The most straightforward is that AnimalInCatOut is a subtype of AnimalInAnimalOut, because wherever you return an Animal you could also return a Cat as it is its subtype. The same thing applies to CatInCatOut  and CatInAnimalOut.

  Might be a bit trickier to see that AnimalInAnimalOut is a subtype of CatInAnimalOut. The first one can replace the second one on call-site without breaking type safety, because the caller expects something the needs a Cat , and so they must provide an argument that is subtype of Cat, and this always satisfy Animal. Of course the same applies to AnimalInCatOut and CatInCatOut.

AnimalInCatOut and CatInAnimalOut combine both treats of the upper two cases; both the former's input a supertype, the output the subtype of the latter. So AnimalInCatOut is a subtype of CatInAnimalOut. Lastly, CatInCatOut cannot be a subtype of AnimalInAnimalOut as it breaks the contract of its input parameter, vice versa it breaks the output, so there is no relation between them. To sum things up:

Now, onto generics. Let's make one generic class by depending on two types: In and Out.

The same rules applies here as well:

• Function<Animal, Cat> is a subtype of Function<Animal, Animal> and Function<Cat, Cat> is a subtype of Function<Cat, Animal>. Here the subtype relation's direction of the dependent type Function is the same as the parameter Out, making Out a covariant type parameter.
• Function<Animal, Animal> is a subtype of Function<Cat, Animal>, Function<Animal, Cat> is a subtype of Function<Cat, Cat>. Here the subtype relation's direction of the dependent type Function is opposite to the direction of the subtype relation between the parameter In making In a contravariant type parameter.
• Making use of both parameters' variance, Function<Animal, Cat> is a subtype of Function<Cat, Animal>.

In some languages, like Scala, variance is declared class level, using + for covariant, - for contravariant parameters.

In Java however, the developer has a cumbersome job: the language does not known class level variance. Instead one has to specify the type variables' variance every place they are used: ? extends T for covariant, ? super T for contravariant positions. If you miss to specify the variance at a place, you lose the information for consequent uses.

Hope this clarified things a bit about type variance. Cheers!