Question

我在编写一个名为 head 的函数时遇到了问题，它基本上将头元素替换为另一个用于调用它的List：

List(1,2,3,4).head(4) // List(4,2,3,4)

代码显然没用，我只是想和Scala一起玩。这是代码：

sealed trait List[+A]{
  def tail():List[A]
  def head[A](x:A):List[A]
}

object Nil extends List[Nothing]{
  def tail() = throw new Exception("Nil couldn't has tail")
  def head[A](x:A): List[A] = List(x)
}

case class Cons[+A](x :A, xs: List[A]) extends List[A]{
  def tail():List[A] = xs
  def head[A](a:A): List[A] = Cons(a,xs)
}

object List{
  def apply[A](as:A*):List[A] = {
    if (as.isEmpty) Nil
    else Cons(as.head,apply(as.tail: _*))
  }
}

Cons(1,Cons(2,Nil)) == List(1,2)
Cons(1,Cons(2,Cons(3,Cons(4,Nil)))).tail()
List(1,2,3,4,5,6,7).tail()
List(1,2,3,4).head(4)

它没有编译，我有这个错误：

Error:(11, 39) type mismatch;
found   : A$A318.this.List[A(in class Cons)]
required: A$A318.this.List[A(in method head)]
 def head[A](a:A): List[A] = Cons(a,xs)

你能解释一下原因吗？

问候。

Answer 1

Your problem is that your head method is taking another type A, therefore inside that scope the compiler takes those As as different, i.e., the A defined in the trait is shadowed by the A in head[A].

Also, your head method is taking a covariant element of type A in a contravariant position, so you can't define head as such.

What you can do is defining your head as:

def head[B >: A](x: B): List[B]

Hence, you get:

object S {
    sealed trait List[+A] {
        def tail(): List[A]
        def head[B >: A](x: B): List[B]
    }

    case object Nil extends List[Nothing] {
        def tail() = throw new Exception("Nil doesn't have a tail")
        def head[B >: Nothing](x: B): List[B] = Cons(x, Nil)
    }

    case class Cons[+A](x: A, xs: List[A]) extends List[A] {
      def tail(): List[A] = xs
      def head[B >: A](a: B): List[B] = Cons(a, xs)
    }

    object List {
      def apply[A](as: A*): List[A] = {
        if (as.isEmpty) Nil
        else Cons(as.head, apply(as.tail: _*))
      }
    }
}

Testing this on the REPL:

scala> :load test.scala
Loading test.scala...
defined object S

scala> import S._
import S._

scala> Nil.head(1)
res0: S.List[Int] = Cons(1,Nil)

scala> Cons(1, Nil).head(4)
res1: S.List[Int] = Cons(4,Nil)

Answer 2

您的方法头不需要类型参数，因为您已经在类定义中定义了一个类型参数，这正是您希望头部接收的类型（如果您希望头部创建相同类型的原始列表））。

你得到的错误是因为A是方法头的上下文中的两种不同类型（方法中的一种，另一种来自类）。

头部的定义应该是：

def head(x:A):List[A]

带有类型参数的Scala List

2 个答案: