Scala中的协方差逆变误差

从Wikipedia开始的协变类型的定义：

Within the type system of a programming language,a typing rule or a
type constructor is:

• covariant if it preserves the ordering of types (≤),which orders types from more specific to more generic;
• contravariant if it reverses this ordering;

考虑以下：

class Fruit
class Banana extends Fruit

和

Banana <: Fruit
A[Banana] :> A[Fruit] // Because A is contravariant in type T
A[A[Banana]] <: A[A[Fruit]] // Again because A is contravariant in type T

最后一个语句意味着A [A [T]]是协变的,因为它保留了类型的排序,从更具体到更通用.

所以有可能做到：

scala> type AA[+T] = A[A[T]]
defined type alias AA

scala> type AAA[-T] = A[A[A[T]]]
defined type alias AAA

但以下将导致错误：

scala> type AA[-T] = A[A[T]]
<console>:15: error: contravariant type T occurs in covariant position in type [-T]A[A[T]] of type AA
       type AA[-T] = A[A[T]]
            ^

scala> type AAA[+T] = A[A[A[T]]]
<console>:15: error: covariant type T occurs in contravariant position in type [+T]A[A[A[T]]] of type AAA
       type AAA[+T] = A[A[A[T]]]
            ^

最后返回到原始问题,因为基类A [A [T]]在类型T中通过构造是协变的,所以在类B定义中存在相同的方差规则违反.

为了理解它被禁止的原因,让我们假设它是可能的：

class B[-T] extends A[A[T]]

在这种情况下,我们得到：

B[Fruit] <: B[Banana] <: A[A[Banana]]

所以

val fruits: B[Fruit] = new B[Fruit]
val bananas1: B[Banana] = fruits
val bananas2: A[A[Banana]] = bananas1

现在我们有一个A [a [香蕉]]的值香蕉2,它指出A [A [水果]]的一个例子,它违反了类型安全,因为A [A [香蕉]]&lt ;: A [A [水果] ].