Scala 中的简明条件运算符

Concise conditional operator in Scala

让我演示一下我对条件运算符的期望:

import scalaz._
import Scalaz._
import util.Random

trait Card

class BigCard extends Card

class SmallCard extends Card

object Main extends App {
  def printCard(card: Card) {println(card.getClass.getSimpleName)}

  // Line bellow works fine, but I would prefer less verbose conditional operator (like ?: from JavaScript).
  (if (Random.nextBoolean) new BigCard else new SmallCard) |> printCard

  // I thought ScalaZ's ?| was meant to be an alternative to if. But following statement fails to compile. 
  (Random.nextBoolean ? new BigCard | new SmallCard) |> printCard
}

结果:

[error] xxx/Main.scala:15: type mismatch;
[error]  found   : SmallCard
[error]  required: BigCard
[error]   (Random.nextBoolean ? new BigCard | new SmallCard) |> printCard
[error]                                       ^
[error] one error found
[error] (compile:compileIncremental) Compilation failed

是否有 ScalaZ 的替代方案?|支持子类的运算符(不确定术语,还是类型扩展)?

我正在寻找一个简洁的条件运算符(因此手动添加类型是不可能的,它最终会比 if 更长更丑陋)。是否可以轻松添加(如没有自定义宏)或某些库是否提供此类运算符?

是的,这很容易(implicit class 不能在顶层声明,将其放入 object 并导入其内容):

class Cond[A](x: Boolean, value: => A) {
  def |[B >: A](other: => B) = if (x) value else other
}

implicit class CondOp(x: Boolean) {
  def ?[A](y: => A) = new Cond(x, y)
}

Scalaz 显然定义了 def |(other: A)

缺点(大概也是 Scalaz 不这样做的原因)是它还针对不相关的类型进行编译,因为在 Scala 中 any 两种类型具有共同的超类型:Random.nextBoolean ? 1 | ""。此外,对于 x ? 1 | 1.0,您不会像 if 那样扩大到 Double;它 returns AnyVal 而不是。