I am confused by the generic subtyping.
In Java, if type A
is a subtype of B
, generic type C<A>
and C<B>
are invariant. For instance, ArrayList<Base>
is not a subtype of ArrayList<Derived>
.
However, in Scala, generic type C<A>
and C<B>
are covariant if type A
is a subtype of B
. So what's the property of generic class in Scala has but not in Java?
Most Scala generic classes are collections, such as the immutable List, Queue, Set, Map, or their mutable equivalents, and Stack. Collections are containers of zero or more objects. We also have generic containers that aren't so obvious at first.
Generics means parameterized types. The idea is to allow type (Integer, String, … etc., and user-defined types) to be a parameter to methods, classes, and interfaces. Using Generics, it is possible to create classes that work with different data types.
To use a generic class, put the type in the square brackets in place of A . The instance stack can only take Ints. However, if the type argument had subtypes, those could be passed in: Scala 2.
In a nutshell, generics enable types (classes and interfaces) to be parameters when defining classes, interfaces and methods. Much like the more familiar formal parameters used in method declarations, type parameters provide a way for you to re-use the same code with different inputs.
Firstly note that variance is a property of generic type parameters, not of the parameterized types themselves.
Secondly: you are wrong about scala - type parameters are invariant by default. Let us investigate!
Java has use-site variance annotations. That is, you can declare methods like this:
boolean addAll(Collection<? extends T> c);
There is, however, one form of "parameterized type" (I use the term loosely) in which the type parameters are covariant: Java Arrays! (This is actually insane because Arrays are mutable and hence it is easy to circumvent the type system). Consider the following:
public static void subvert(Object[] arr) { arr[0] = "Oh Noes!"; }
And then:
Integer[] arr = new Integer[1];
subvert(arr); //this call is allowed as arrays are covariant
Integer i = arr[0];
A good interview question this one: what happens?
In Scala, you have declaration-site variance. That is, the variance of a type parameter is declared alongside the parameter (using the annotations +
and -
):
trait Function1[-I, +O]
This says that the trait Function1
has two type parameters, I
and O
which are contra- and co-variant respectively. If no +/-
annotation is declared, then the type parameter is invariant. For example, Set is invariant in its type parameter:
scala> def foo(set: Set[Any]) = ()
foo: (set: Set[Any])Unit
scala> Set(1)
res4: scala.collection.immutable.Set[Int] = Set(1)
scala> foo(res4)
<console>:10: error: type mismatch;
found : scala.collection.immutable.Set[Int]
required: Set[Any]
Note: Int <: Any, but trait Set is invariant in type A.
You may wish to investigate a wildcard type such as `_ <: Any`. (SLS 3.2.10)
foo(res4)
^
List is however, declared as being covariant:
scala> def bar(list: List[Any]) = ()
bar: (list: List[Any])Unit
scala> List(1)
res6: List[Int] = List(1)
scala> bar(res6)
Another way of demonstrating this is to ask the compiler directly for subtype-evidence:
scala> class Cov[+A]
defined class Cov
scala> implicitly[Cov[Int] <:< Cov[Any]]
res8: <:<[Cov[Int],Cov[Any]] = <function1>
But with an invariant type parameter
scala> class Inv[A]
defined class Inv
scala> implicitly[Inv[Int] <:< Inv[Any]]
<console>:9: error: Cannot prove that Inv[Int] <:< Inv[Any].
implicitly[Inv[Int] <:< Inv[Any]]
^
Lastly, contravariance:
scala> class Con[-A]
defined class Con
scala> implicitly[Con[Any] <:< Con[Int]]
res10: <:<[Con[Any],Con[Int]] = <function1>
See also identifier <:<
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