虽然这样编译你不能用坏类型实例化这个类型:
class Wololo extends TypedTrait { type Tpe = Type.Type2.type } // bad type
val ax: Test.Aux[Wololo] = new Test { type TptTrait = Wololo }
这失败了
cmd1.sc:24: incompatible type in overriding
type TptTrait <: ammonite.$sess.cmd1.TypedTrait.Aux[ammonite.$sess.cmd1.Type.Type1.type] (defined in trait Test);
found : ammonite.$sess.cmd1.Wololo
required: <: ammonite.$sess.cmd1.TypedTrait.Aux[ammonite.$sess.cmd1.Type.Type1.type]
(which expands to) <: ammonite.$sess.cmd1.TypedTrait{type Tpe = ammonite.$sess.cmd1.Type.Type1.type}
val ax: Test.Aux[Wololo] = new Test { type TptTrait = Wololo }
^
Compilation Failed
同时
class Wololo extends TypedTrait { type Tpe = Type.Type1.type } // good type
val ax: Test.Aux[Wololo] = new Test { type TptTrait = Wololo }
成功
ax: Test.Aux[Wololo] = ammonite.$sess.cmd1$$anon$1@78de58ea
所以我们可以得出结论,这里的边界结合了,但在您实际尝试创建一个违反约束的实例之前不会失败。可能这不会触发规范让它立即失败的任何情况(毕竟你可以创建一个有效的实例!),所以检查/证明被推迟到你尝试创建一个实际的实例。
sealed trait
(因为s ,我在 Ammonite 中将其作为一个整体进行了评估)
@ {
sealed trait Type
object Type {
case object Type1 extends Type
case object Type2 extends Type
}
sealed trait TypedTrait {
type Tpe <: Type
}
object TypedTrait {
type Aux[T <: Type] = TypedTrait{ type Tpe = T }
}
sealed trait Test {
//Bounded with TypedTrait.Aux[Type.Type1.type]
type TptTrait <: TypedTrait.Aux[Type.Type1.type]
}
object Test {
type Aux[T <: TypedTrait] = Test { type TptTrait = T }
}
class Wololo extends TypedTrait { type Tpe = Type.Type1.type }
val ax: Test.Aux[Wololo] = new Test { type TptTrait = Wololo }
}
defined trait Type
defined object Type
defined trait TypedTrait
defined object TypedTrait
defined trait Test
defined object Test
defined class Wololo
ax: Test.Aux[Wololo] = ammonite.$sess.cmd1$$anon$1@78de58ea