scala - 强制类型差异

Question

在 Scala 中，我可以在编译时强制执行类型相等。例如:

case class Foo[A,B]( a: A, b: B )( implicit ev: A =:= B )

scala> Foo( 1, 2 )
res3: Foo[Int,Int] = Foo(1,2)

scala> Foo( 1, "2" )
<console>:10: error: Cannot prove that Int =:= java.lang.String.

有没有办法强制类型 A 和类型 B 应该不同？

Answer 1

我有一个更简单的解决方案，它也利用了歧义，

trait =!=[A, B]

implicit def neq[A, B] : A =!= B = null

// This pair excludes the A =:= B case
implicit def neqAmbig1[A] : A =!= A = null
implicit def neqAmbig2[A] : A =!= A = null

原始用例，

case class Foo[A,B](a : A, b : B)(implicit ev: A =!= B)
new Foo(1, "1")
new Foo("foo", Some("foo"))

// These don't compile
// new Foo(1, 1)
// new Foo("foo", "foo")
// new Foo(Some("foo"), Some("foo"))

更新

我们可以将其与我的 "magical typesystem tricks" 联系起来(感谢@jpp；-)如下，

type ¬[T] = T => Nothing
implicit def neg[T, U](t : T)(implicit ev : T =!= U) : ¬[U] = null

def notString[T <% ¬[String]](t : T) = t

示例 REPL session ，

scala> val ns1 = notString(1)
ns1: Int = 1

scala> val ns2 = notString(1.0)
ns2: Double = 1.0

scala> val ns3 = notString(Some("foo"))
ns3: Some[java.lang.String] = Some(foo)

scala> val ns4 = notString("foo")
<console>:14: error: No implicit view available from 
  java.lang.String => (String) => Nothing.
       val ns4 = notString2("foo")
                            ^