斯卡拉：動態訪問同伴從子類

Question

我想與同伴幾個類，以便從一個類的實例，以及從同伴對象訪問共享值的對象，當我沒有一個實例對象。

最後我想解析字符串和圓創建匹配的類的實例，也是另一種方式，創建從實例的字符串。

下面的代碼工作，但似乎過於複雜：

abstract class Person(val age: Int = 0) { 
    val pronoun: String 
    override def toString = pronoun + " is " + age + " years old" 
} 

class Man(override val age: Int) extends Person { 
    val pronoun = Man.pronoun 
} 

class Woman(override val age: Int) extends Person { 
    val pronoun = Woman.pronoun 
} 

object Man extends Person { 
    val pronoun = "he" 
} 

object Woman extends Person { 
    val pronoun = "she" 
} 

object Person { 
    def fromString(pronoun: String, age: Int): Option[Person] = { 
     pronoun match { 
      case Man.pronoun => Some(new Man(age)) 
      case Woman.pronoun => Some(new Woman(age)) 
      case _ => None 
     } 
    } 
} 

// Man.pronoun 
// -> should return "he" 

// Woman.pronoun 
// -> should return "she" 

// new Man(30).toString 
// -> should return "he is 30 years old" 

// Person.fromString("she", 20) 
// -> should return a Some(Woman) with age = 20 

所以，對於這個工作，我爲每個子類的類和同伴對象和手動引用來自同伴對象pronoun（例如val pronoun = Man.pronoun）。

邊注：在抽象類設置的默認值age = 0使事情進一步下跌更容易一點，但它似乎沒有吧？

我使用特點等，但都沒有成功嘗試。有沒有更好的方式來實現這個重複性較低的代碼？

Answer 1

鑑於這個問題，我會做它簡化：

case class Person(pronoun: String, age: Int) { 
    override def toString = s"$pronoun is $age years old" 
} 

，並應解決這個問題，但我猜你提出的問題是一個比較複雜的一個的簡化，所以我的解決方案成爲：

trait Person { 
    protected def age: Int 
    protected def pronoun: String 
    override def toString = s"$pronoun is $age years old" 
} 

case class Man(override val age: Int) extends Person { 
    override protected def pronoun: String = Person.SHE 
} 
case class Woman(override val age: Int) extends Person { 
    override protected def pronoun: String = Person.HE 
} 

object Person { 

    val HE = "he" 
    val SHE = "she" 

    def parse(pronoun: String, age: Int): Option[Person] = { 
    pronoun match { 
     case HE => Some(Man(age)) 
     case SHE => Some(Woman(age)) 
     case _ => throw new IllegalArgumentException(s"pronoun $pronoun was not recognized") 
    } 
    } 

} 

這對我來說看起來並不糟糕，解決了這個問題並且非常可讀。說了這麼多，你的代碼看起來不錯（默認是0）。這看起來很複雜，但我花了2秒才明白自己在做什麼，而且可讀性近來非常重要。

希望這會幫助你。問候！

Answer 2

什麼是你必須遵守的約束？例如。你必須能夠調用Man.pronoun？因爲一個選擇是簡單地創建在Person對象ManPronoun和WomanPronoun領域和刪除Man和Woman同伴對象。另外，什麼是0默認年齡值的點？您可以簡單地將年齡傳遞給Person。

abstract class Person(val age: Int) { 
    val pronoun: String 
    override def toString = pronoun + " is " + age + " years old" 
} 

class Man(override val age: Int) extends Person(age) { 
    val pronoun = Person.ManPronoun 
} 

class Woman(override val age: Int) extends Person(age) { 
    val pronoun = Person.WomanPronoun 
} 

object Person { 

    val ManPronoun = "he" 
    val WomanPronoun = "she" 

    def fromString(pronoun: String, age: Int): Option[Person] = { 
    pronoun match { 
     case ManPronoun => Some(new Man(age)) 
     case WomanPronoun => Some(new Woman(age)) 
     case _   => None 
    } 
    } 
} 

如果你必須有內部Man和Woman對象的代名詞場，好，那你必須有Man和Woman對象。 :)