Scala 函数子类型

Question

type Fun1 = { val a: A } => { val b: B }

type Fun2 = { val b: B } => { val a: A }

我想知道的是上述类型的父类(super class)型。

Fun1 <: 普通

Fun2 <: 普通

Answer 1

由于函数的输入类型是逆变，而返回类型是协变，所以公共(public)输入类型是“最接近”的公共(public)子类型 A 和 B(例如 A with B)，并且公共(public)返回类型是 A 和 B 的“最接近”父类(super class)型(例如 Any ), 所以如果我们对 A 和 B 一无所知，答案是:

A with B => Any

例如:

scala> class A
defined class A

scala> class B
defined class B

scala> type Fun1 = A => B
defined type alias Fun1

scala> type Fun2 = B => A
defined type alias Fun2

scala> val f1: Fun1 = (x: A) => new B()
f1: A => B = <function1>

scala> val f2: Fun2 = (x: B) => new A()
f2: B => A = <function1>

// both f1 and f2 can be assigned into a value of type A with B => Any
scala> val f3: A with B => Any = f1
f3: A with B => Any = <function1>

scala> val f4: A with B => Any = f2
f4: A with B => Any = <function1>