scala - 为什么适当的类型 Any 和 Nothing 在类型不同时适合类型构造函数形状 F[_] ？

Question

考虑以下采用 * -> * 类型参数的方法

def g[F[_]] = ???

为什么下面不是语法错误

g[Any]       // ok
g[Nothing]   // ok

自从

scala> :kind -v Any
Any's kind is A
*
This is a proper type.

scala> :kind -v Nothing
Nothing's kind is A
*
This is a proper type.

所以 Any 和 Nothing 应该是错误的形状？

Answer 1

Scala 规范引述:

For every type constructor 𝑇 (with any number of type parameters), scala.Nothing <: 𝑇 <: scala.Any.

https://scala-lang.org/files/archive/spec/2.13/03-types.html#conformance

Say the type parameters have lower bounds 𝐿1,…,𝐿𝑛 and upper bounds 𝑈1,…,𝑈𝑛. The parameterized type is well-formed if each actual type parameter conforms to its bounds, i.e. 𝜎𝐿𝑖<:𝑇𝑖<:𝜎𝑈𝑖 where 𝜎 is the substitution [𝑎1:=𝑇1,…,𝑎𝑛:=𝑇𝑛].

https://scala-lang.org/files/archive/spec/2.13/03-types.html#parameterized-types

A polymorphic method type is denoted internally as [tps]𝑇 where [tps] is a type parameter section [𝑎1 >: 𝐿1 <: 𝑈1,…,𝑎𝑛 >: 𝐿𝑛 <: 𝑈𝑛] for some 𝑛≥0 and 𝑇 is a (value or method) type. This type represents named methods that take type arguments 𝑆1,…,𝑆𝑛 which conform to the lower bounds 𝐿1,…,𝐿𝑛 and the upper bounds 𝑈1,…,𝑈𝑛 and that yield results of type 𝑇.

https://scala-lang.org/files/archive/spec/2.13/03-types.html#polymorphic-method-types

所以自 Any和 Nothing符合F[_]的上下界(即Any和Nothing对应)，g[Any]和 g[Nothing]是合法的。