[英]Scala - object extends abstract class and takes parameters
//文件Animal.scala
abstract class Animal {
val name: String
def getSomething(tClass: TypeClass): String = {
tClass.tName.split('.').lift(0)
}
def apply(tClass: TypeClass): SomeOtherClassType = {
// something...
}
// File: DogSpike ,用於某些特定情況(覆蓋基類val)
object DogSpike extends Animal {
override val name: String = "Spike"
}
然后,此通話有效(通話適用)
myTransformation(() => DogSpike(this))
現在,我想創建一個更通用的對象,該對象可以傳遞參數,但是我無法傳遞參數。
從Animal創建派生對象將采用一個參數,並且能夠使用apply調用將起作用
object TheDog(name: String) extends Animal {
override val name: String = "Spike"
//...
}
不知道如何隱式調用TheDog
對象的Animal.apply,我可以在其中傳遞參數(名稱)
myTransformation(() => TheDog(this))
// also this seems to be incorrect "*Wrong top statement declaration*"
object TheDog(val n: String) extends Animal {
override val name: String = n
//...
}
從*Wrong top statement declaration*
(我只能理解您問題的這一部分)-您不能在對象中使用構造函數,因為object
是單例,因此您應該使用案例類(ADT):
final case class TheDog(name: String) extends Animal
scala>TheDog("Spike")
res2_2: TheDog = TheDog("Spike")
val
和伴侶的對象與apply
的情況下,為類自動添加,所以你並不需要定義自己的自己的apply
在Animal
。 case class TheDog(val name: String)
與case class TheDog(name: String)
。
我還使用trait
而不是抽象類:
trait Animal {
val name: String
def getSomething: String = {
"Dog: " + name
}
}
我不了解您的TypeClass
類型,但是如果您確實想要類型類:
trait Animal {
def name: String
}
final case class TheDog(name: String) extends Animal
final case class TheCat(name: String) extends Animal
implicit class RichDog(dog: TheDog){
def getSomething: String = {
"Dog" + dog.name
}
}
implicit class RichCat(cat: TheCat){
def getSomething: String = {
"Cat: " + cat.name
}
}
scala> TheDog("Spike").getSomething
res4_5: String = "DogSpike"
scala> TheCat("Tom").getSomething
res4_6: String = "Cat: Tom"
關於“隱式”調用apply
,我不知道為什么有人需要這個,但是:
trait AnimalFactory[A <: Animal] {
def apply(name: String)(implicit constructor: String => A) = constructor(name)
}
object TheeeDog extends AnimalFactory[TheDog]
implicit def createDog(name: String) = TheDog(name)
TheeeDog("Spike")
當然,您必須提供createDog
並使它對客戶端可見,但是如果您僅可以使用ADT並在伴隨對象中定義其他必需的apply
,那實際上是沒有意義的:
case class TheMouse(name: String)
object TheMouse{
def apply(isJerry: Boolean): TheMouse = if (isJerry) TheMouse("Jerry") else TheMouse("NotJerry")
}
TheMouse(true)
如果要向構造函數添加一些參數,只需添加它:
class AnimalFactory(clazz: SomeClass){
def doSomething = clazz.name
def apply(name: String)
}
val dogFactory = new AnimalFactory(dogClassDescriptor)
val catFactory = new AnimalFactory(catClassDescriptor)
dogFactory("Spike")
catFactory("Tom")
您甚至可以為工廠創建工廠(我不建議-該解決方案看起來已經很復雜了):
object AnimalFactory{ //please don't use classes for that - avoiding `new` is not their purpose
def apply(clazz: SomeClass) = new AnimalFactory(clazz)
}
val dogFactory = AnimalFactory(dogClassDescriptor)
//or even `val spike = AnimalFactory(dogClassDescriptor)("Spike")`
不過還是有什么,如果你可以只提供基本的點clazz
無論是作為一個成員或只是一個包裝:
final case class ClazzWrapper[T <: Animal](clazz: SomeClass, animal: T)
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