可能重复:
Enforce type difference
由于在scala =:=
中存在强制执行相等性的通用类型约束,是否存在对类型强制执行“不等于”的约束?基本上!=
但是对于类型?
修改
下面的评论指向现有的Q&A,答案似乎是(1)不,它不在标准库中(2)是的,可以定义一个。
所以我会修改我的问题,因为在看到答案后我发现了一个想法。
鉴于现有解决方案:
sealed class =!=[A,B]
trait LowerPriorityImplicits {
implicit def equal[A]: =!=[A, A] = sys.error("should not be called")
}
object =!= extends LowerPriorityImplicits {
implicit def nequal[A,B](implicit same: A =:= B = null): =!=[A,B] =
if (same != null) sys.error("should not be called explicitly with same type")
else new =!=[A,B]
}
case class Foo[A,B](a: A, b: B)(implicit e: A =!= B)
如果A <: B
或A >: B
,A =!= B
仍然如此吗?如果没有,是否可以修改解决方案,以便A =!= B
如果不是A <: B
或A >: B
?
答案 0 :(得分:11)
shapeless使用与严格类型不等式相同的隐式歧义技巧来定义类型运算符A <:!< B
(意思是A
不是B
的子类型)
trait <:!<[A, B]
implicit def nsub[A, B] : A <:!< B = new <:!<[A, B] {}
implicit def nsubAmbig1[A, B >: A] : A <:!< B = sys.error("Unexpected call")
implicit def nsubAmbig2[A, B >: A] : A <:!< B = sys.error("Unexpected call")
示例REPL会话,
scala> import shapeless.TypeOperators._
import shapeless.TypeOperators._
scala> implicitly[Int <:!< String]
res0: shapeless.TypeOperators.<:!<[Int,String] =
shapeless.TypeOperators$$anon$2@200fde5c
scala> implicitly[Int <:!< Int]
<console>:11: error: ambiguous implicit values:
both method nsubAmbig1 in object TypeOperators of type
[A, B >: A]=> shapeless.TypeOperators.<:!<[A,B]
and method nsubAmbig2 in object TypeOperators of type
[A, B >: A]=> shapeless.TypeOperators.<:!<[A,B]
match expected type shapeless.TypeOperators.<:!<[Int,Int]
implicitly[Int <:!< Int]
^
scala> class Foo ; class Bar extends Foo
defined class Foo
defined class Bar
scala> implicitly[Foo <:!< Bar]
res2: shapeless.TypeOperators.<:!<[Foo,Bar] =
shapeless.TypeOperators$$anon$2@871f548
scala> implicitly[Bar <:!< Foo]
<console>:13: error: ambiguous implicit values:
both method nsubAmbig1 in object TypeOperators of type
[A, B >: A]=> shapeless.TypeOperators.<:!<[A,B]
and method nsubAmbig2 in object TypeOperators of type
[A, B >: A]=> shapeless.TypeOperators.<:!<[A,B]
match expected type shapeless.TypeOperators.<:!<[Bar,Foo]
implicitly[Bar <:!< Foo]
^