为什么创建元组的显式语法只允许AnyRefs作为类型注释？

Question

此代码适用于：

scala> val x = ""
x: java.lang.String = ""

scala> Tuple2[x.type, x.type](x,x)
res5: (x.type, x.type) = ("","")

这张没有：

scala> val y = 0
y: Int = 0

scala> Tuple2[y.type, y.type](y,y)
<console>:9: error: type mismatch;
 found   : y.type (with underlying type Int)
 required: AnyRef
Note: an implicit exists from scala.Int => java.lang.Integer, but
methods inherited from Object are rendered ambiguous.  This is to avoid
a blanket implicit which would convert any scala.Int to any AnyRef.
You may wish to use a type ascription: `x: java.lang.Integer`.
              Tuple2[y.type, y.type](y,y)
                     ^

就像这个一样：

scala> val z = ()
z: Unit = ()

scala> Tuple2[z.type, z.type](z,z)
<console>:9: error: type mismatch;
 found   : z.type (with underlying type Unit)
 required: AnyRef
Note: Unit is not implicitly converted to AnyRef.  You can safely
pattern match `x: AnyRef` or cast `x.asInstanceOf[AnyRef]` to do so.
              Tuple2[z.type, z.type](z,z)
                     ^

语言规范说

单例类型为p.type形式，其中p是指向预期符合scala.AnyRef (§6.1)的值的路径。

这背后的理由是什么，是否有必要像最近0.getClass那样取消这种限制？

Answer 1

正如您在Scala类层次结构 Int上看到的那样，Unit、Boolean (和其他类)不是AnyRef的子类，因为它们被转换为java int、void、boolean等等，女巫不是Object的子类。