重寫方法簽名，以避免調用.asInstanceOf客戶端代碼

Question

在這一段代碼下面我明白爲什麼我必須按順序調用asInstanctOf能夠調用bazMethod從類Baz：

class Bar[BR] 
class Baz extends Bar[Int] { 
    def bazMethod: Unit = {} 
} 

def map[A, B](args: Seq[Bar[A]])(f: Bar[A] => B): Seq[B] = args.map(f) 

//Client Code 
val bazes = Seq(new Baz) 
map(bazes)(b => b.asInstanceOf[Baz].bazMethod) 

但是什麼我想知道的是是否有方法來重寫此客戶端代碼不必使用asInstanceOf。 只有一個條件：我需要參考Bar[A]在方法中以某種方式提供給我。想象這是一個更復雜的方法，其中的東西需要根據A究竟是什麼來完成。

Answer 1

如果你有map接受T <: Bar[_]那麼它應該工作：

scala> :pa 
// Entering paste mode (ctrl-D to finish) 

class Bar[BR] 
class Baz extends Bar[Int] { 
    def bazMethod: Unit = {} 
} 

def map[B, T <: Bar[_]](args: Seq[T])(method: T => B): Seq[B] = args.map(method) 

// Exiting paste mode, now interpreting. 

defined class Bar 
defined class Baz 
map: [B, T <: Bar[_]](args: Seq[T])(method: T => B)Seq[B] 

scala> val bazes = Seq(new Baz) 
bazes: Seq[Baz] = List(Baz@225129c) 

scala> map(bazes)(b => b.bazMethod) 
res0: Seq[Unit] = List(()) 

否則，@碧玉-M所指出的，如果你想訪問類型參數A，你會做這樣的事情：

scala> :pa 
// Entering paste mode (ctrl-D to finish) 

class Bar[BR] 
class Baz extends Bar[Int] { 
    def bazMethod: Unit = {} 
} 

def map[B, T <: Bar[A], A](args: Seq[T with Bar[A]])(method: T => B): Seq[B] = args.map(method) 

// Exiting paste mode, now interpreting. 

defined class Bar 
defined class Baz 
map: [B, T <: Bar[A], A](args: Seq[T with Bar[A]])(method: T => B)Seq[B] 

scala> val bazes = Seq(new Baz) 
bazes: Seq[Baz] = List(Baz@341814d3) 

scala> map(bazes)(b => b.bazMethod) 
res0: Seq[Unit] = List(())