三元运算符输入

Question

我实现了一个三元运算符如Java的<condition> ? <if true> : <if false>,替代/的:,因为:不是一个有效的标识符:

case class Ternary[T](val o: Option[T]) {
  def / (f: => T) = o getOrElse f
}

implicit def boolToTernary(cond: Boolean) = new {
  def ? [T](f: => T) = if(cond) Ternary(Some(f)) 
                        else    Ternary[T](None)
}

它一般工作正常,例如

scala> (1 > 2) ? "hi" / "abc"
res9: java.lang.String = abc

但在以下情况下跌倒:

scala> (1 > 2) ? 5 / 6.0
<console>:33: error: type mismatch;
 found   : Double(6.0)
 required: Int
       (1 > 2) ? 5 / 6.0
                     ^

我是否可以对类型进行调整以使其像内置一样工作if (1 > 2) 5 else 6.0？我搜索了类似的解决方案,我发现的实现都表现出相同的行为.

Answer 1

您可以做的一件事是将您的定义更改/为:

def /[U >: T](f: => U) = o getOrElse f

它不像普通的那样if(推断类型会是这样Double) - 你得到了AnyVal,但这已经是一个改进,只需对代码进行非常小的修改.

更新:我认为我已经找到了一种(稍微复杂一点)的方法,使其表现得更像普通的if(例如,在这种情况下,推断a Double).试试这段代码:

implicit def boolToTernary(cond: Boolean) = new {
  def ?[T](f: => T) = if (cond) Ternary(Some(f))
  else Ternary[T](None)
}

case class Ternary[T](val o: Option[T]) {
  def /[U, That](f: => U)(implicit conv: BiConverter[T, U, That]): That = o map conv.toThat1 getOrElse (conv toThat2 f)
}

class BiConverter[T, U, That](val toThat1: T => That, val toThat2: U => That)

trait LowPriorityBiConverterImplicits {
  implicit def subtype[A, T <: A, U <: A]: BiConverter[T, U, A] = new BiConverter[T, U, A](identity[T], identity[U])
}

object BiConverter extends LowPriorityBiConverterImplicits {
  implicit def identityConverter[T]: BiConverter[T, T, T] = new BiConverter[T, T, T](identity, identity)
  implicit def firstAsSecond[T, U](implicit conv: T => U): BiConverter[T, U, U] = new BiConverter[T, U, U](conv, identity)
  implicit def secondAsFirst[T, U](implicit conv: U => T): BiConverter[T, U, T] = new BiConverter[T, U, T](identity, conv)
}

然后(一些示例代码):

abstract class Fruit
class Apple extends Fruit
class Banana extends Fruit

def main(args: Array[String]) {
  val int = (1 > 2) ? 5 / 6 // Int is inferred
  val fruit = (1 > 2) ? new Apple / new Banana // Fruit is inferred
  val double1 = (1 > 2) ? 5 / 5.5 // Double is inferred
  val double2 = (1 > 2) ? 5.5 / 5 // Double is inferred
}

Answer 2

这是我的版本,因为我很好奇.我不会真的用这个......

我选择了低优先级运营商.^先绑定,然后绑定|?.我使用了元组是协变的事实来推断结果的类型.

case class TernClause[T](t: T) {
  def ^[U](u: U) = (t, u)
}
case class Tern(b: Boolean) {
  def |?[U](tuple: (U,U)) = if (b) tuple._1 else tuple._2
}

implicit def toTern(b: Boolean): Tern = Tern(b)
implicit def toTernClause[T](t: T): TernClause[T] = TernClause(t)

(1 > 2) |? "hi" ^ "abc"
// java.lang.String = abc

(1 > 2) |? 5 ^ 6.0
// AnyVal{def getClass(): java.lang.Class[_ >: Double with Int <: AnyVal]} = 6.0

显示运算符优先级如何协同工作的另一个示例

3 > 2 |? 5 - 1 ^ 6.0 + 1
// AnyVal{def getClass(): java.lang.Class[_ >: Double with Int <: AnyVal]} = 4

可能需要一些工作来确保未使用的分支不被评估.

只是想法的食物:请注意,在调用函数时类型推断是正确的.一个不太灵活的语法会对你有用吗？这真的很简单......

class BooleanEx(b: Boolean) {
  def ?[U](onTrue: => U, onFalse: => U) = if (b) onTrue else onFalse
}

implicit def toBooleanEx(b: Boolean): BooleanEx = new BooleanEx(b)

class A
class B extends A

(1 > 2) ? ("hi", "abc")
//res0: java.lang.String = abc
(1 > 2) ? (5, 6.0)
//res1: Double = 6.0
(1 > 2) ? (5, 6)
//res2: Int = 6
(1 > 2) ? (new A, new B)
//res3: A = B@1e21540

此外,这在scalaz中可用,但他们将其命名为fold:

import scalaz._
import Scalaz._
(1 > 2) fold (5, 6.0)
//res0: Double = 6.0