# Variance of type parameters in Scala

This is just a quick introduction to one of the features of Scala’s generics. I realised earlier on IRC that they’re probably quite unfamiliar looking to people new to the language, so thought I’d do a quick writeup.

What does the following mean?

  trait Function1[-T1, +R]


It’s saying that the trait Function1 is contravariant in the parameter T1 and covariant in the parameter R.

Err. Eek. Scary words!

Lets try that again.

A Function1[Any, T] is safe to use as a Function1[String, T]. If I can apply f to anything I can certainly apply it to a String. This is contravariance. Similarly, a Function1[T, String] can be quite happily treated as a Function1[String, Any] – if it returns a String, it certainly returns an Any.

So, Foo[+T] means that if S <: T then Foo[S] <: Foo[T]. Foo[-T] means that if S <: T then Foo[T] <: Foo[S] (note the swapped the direction). The default Foo[T], called invariant, is that Foo[S] is not a subtype of Foo[T] unless S == T.

Examples of this sort of behaviour abound. Covariance is more common than contravariance, because immutable collections are almost always covariant in their type parameters. An immutable.List[String] can equally well be treated as an immutable.List[Any] – all the operations are concerned with what values you can get out of the list, so can easily be widened to some supertype.

However, a mutable.List is *not* covariant in its type parameter. You might be familiar with the problems that result from treating it as such from Java. Suppose I have a mutable.List[String], upcast it to a mutable.List[Any] and now do myList += 3. I’ve now added an integer to a list of Strings. Oops! For this reason, mutable objects tend to be invariant in their type parameters.

So, we have three types of type parameter: Covariant, contravariant, invariant. All three crop up and are quite useful.

But there are safe ways to treat mutable objects invariable. Suppose I want someone to pass me an array of Foos, and I have no intention of mutating it. It’s perfectly safe for them to pass me an array of Bars where Bar extends Foo. Can I do this?

Well, this can indeed be done. We could start by doing this:

  def doStuff[T <: Foo](arg : Array[T]) = stuff;


So we introduce a type parameter for the array. Because T will be inferred in most cases, this isn't too painful to use, but it can quickly cause the number of type parameters to explode (and you don't seem to be able to let some type parameters be inferred and some be explicitly provided). Further, we only care about the type parameter in one place. So, let's move it there.

  def doStuff(arg : Array[T forSome { type T <: Foo }]) = stuff;


This uses Scala's existential types to specify that there's an unknown type for which this holds true. This is effectively equivalent to the previous code, but narrows the scope of the type parameter.

The equivalent using Java style wildcards would be:

  def doStuff(arg : Array[? <: Foo ]) = stuff;


But this isn't legal Scala. This is unfortunately a case of Scala being more verbose than the Java equivalent. However, it's not all bad - because of the explicitly named type inside the forSome, you can express more complicated type relationships than wildcards allow for. For example the following:

  def doStuff(arg : Array[T forSome { type T <: Comparable[T]}]) = stuff;


And that's about it for variance in Scala. Hope you found it useful.

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