具有下限的 Scala 协方差 [英] Scala covariance with lower bound

问题描述

我正在看动作手册中的 Scala，它有这个代码

I am looking at scala in action book and it has this code

sealed abstract class Maybe[+A] {
    def isEmpty: Boolean
    def get: A
    def getOrElse[B >: A](default: B): B = {
        if(isEmpty) default else get
    }
}
    final case class Just[A](value: A) extends Maybe[A] {
    def isEmpty = false
    def get = value
}
case object Nil extends Maybe[scala.Nothing] {
    def isEmpty = true
    def get = throw new NoSuchElementException("Nil.get")
}

如果getOrElse的签名定义为def getOrElse(默认:A):A =它无法编译.

If the signature of getOrElse is defined as def getOrElse(default: A): A = It doesnt compile.

作者说下界 B >:A 声明类型参数 B 被约束为类型 A 的某个超类型"

The author states "The lower bound B >: A declares that the type parameter B is constrained to some super type of type A"

然而我似乎能够做到这一点并且它有效

Yet I seem to be able to do this and it works

val j1 = Just(1)                                  
val l1 = j1.getOrElse("fdsf")  //l1  : Any = 1

String 是 Int 的超类型吗?我不明白为什么会这样?就像它回退到参数 1 是类型 A 是类型 Any(它是)而不是类型 Int.

Is String a super type of Int? What am i not understanding as to why this works? Its like its falling back to argument 1 being type A being of type Any (which it is) rather than type Int.

推荐答案

在 Scala 中，方法参数中不能有协变类型.

In Scala you cannot have covariant types in method parameters.

这是因为在方法参数中允许协变类型会破坏类型安全.

This is because allowing covariant types in method parameters breaks type safety.

为了拥有协变类型，您必须使用有界类型:

In order to have a covariant type you must use a bounded type:

getOrElse[B >: A](default: B): B

这表示找到某种类型，B，使其成为A 的超类，并成为方法返回类型.

This says find some type, B, such that it is a superclass of A and that becomes the method return type.

在你的情况下 A 是 Int 并且你传入一个 String.唯一类型 B 是 Int 和 String 的超类是 Any.

In your case A is Int and you pass in a String. The only type B which is a superclass of both Int and String is Any.

在这种情况下，B 变为 Any，因此该方法返回 Any.

In this case B becomes Any so the method returns Any.

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