According to covariance definition:
Q[+B] means that Q can take any class, but if A is a subclass of B,
then Q[A] is considered to be a subclass of Q[B].
Let's see the following example:
trait SomeA
trait SomeB extends SomeA
trait SomeC extends SomeB
case class List1[+B](elements: B*)
val a = List1[SomeA](new SomeA{},new SomeB{})
val b = List1[SomeB](new SomeB{},new SomeC{})
Everything is fine, but I don't see why List1[SomeB] is a subclass of List1[SomeA], or in other words why b is a subclass of a?
Now, List1[SomeB] is subclass of List1[SomeA] which means you can put the first where the later is needed.
scala> case class List1[+B](elements: B*)
scala> val a = List1[SomeA](new SomeA{},new SomeB{})
a: List1[SomeA] = List1(WrappedArray($anon$2@3e48e859, $anon$1@31ddd4a4))
scala> val b = List1[SomeB](new SomeB{},new SomeC{})
b: List1[SomeB] = List1(WrappedArray($anon$2@5e8c34a0, $anon$1@7c1c5936))
scala> val c: List1[SomeA] = b
c: List1[SomeA] = List1(WrappedArray($anon$2@5e8c34a0, $anon$1@7c1c5936))
scala> val c: List1[SomeA] = a
c: List1[SomeA] = List1(WrappedArray($anon$2@3e48e859, $anon$1@31ddd4a4))
If it were invariant that would be not possible, see:
scala> case class List1[B](elements: B*)
defined class List1
scala> val c: List1[SomeA] = b
<console>:16: error: type mismatch;
found : List1[SomeB]
required: List1[SomeA]
Note: SomeB <: SomeA, but class List1 is invariant in type B.
You may wish to define B as +B instead. (SLS 4.5)
val c: List1[SomeA] = b
^
scala> val c: List1[SomeA] = a
c: List1[SomeA] = List1(WrappedArray($anon$2@45acdd11, $anon$1@3f0d6038))
Since all the elements in List1[SomeB] are subtype of SomeA (as SomeB extends SomeA), you can simply pass List1[SomeB] where List1[SomeA] is expected.
