Question

此代码有效：

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

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

这个没有：

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)
                     ^

以及这一个：

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

单例类型的形式为p.type，其中p是指向的路径期望符合（§6.1）scala.AnyRef。
的值

这背后的理由是什么，如果最近发生在0.getClass的情况下解除限制是否有意义？

Answer 1

正如您在Scala class hierarchy Int上看到的那样，Unit，Boolean（和其他人）不是AnyRef的子类，因为它们被翻译成java {{1} }，int，void等等，女巫不是boolean的子类。

为什么创建元组的显式语法只允许AnyRefs作为类型注释？

1 个答案: