如何讓特徵處理字符串和列表？

Question

我想有一個適用於String和List[_]的特徵。那就是：

trait Mixer[A] { 
    def mix(a: A, b: A): A = b ++ a ++ b 
    def isMix(abab: A, b: A) = 
    abab.endsWith(b) && abab.startsWith(b) && abab.length > 2*b.length 
} 

object MixerString extends Mixer[String] 
object MixerListInt extends Mixer[List[Int]] 

失敗的嘗試

trait Mixer[C, A] { 
    implicit def toSeqLike(in: A): collection.SeqLike[C, A] 
    implicit val bf: collection.generic.CanBuildFrom[A, C, A] 
    def mix(a: A, b: A): A = b ++ a ++ b 
    def isMix(abab: A, b: A) = 
    abab.endsWith(b) && abab.startsWith(b) && abab.length > 2*b.length 
} 

它抱怨的方法isMix有：

error: type mismatch;
found : A
required: scala.collection.GenSeq[?]
> abab.endsWith(b) && abab.startsWith(b) && abab.length > b.length
> ^

問題

無論我怎麼解決這最後一個錯誤，或者我該如何做一些簡單的事情？

編輯：完整的解決方案（感謝@Frederico）

import collection.generic.CanBuildFrom 
import collection.SeqLike 
trait Mixer[C, A] { 
    implicit def toSeqLike(in: A): SeqLike[C, A] 
    implicit val bf: CanBuildFrom[A, C, A] 
    def mix(a: A, b: A): A = b ++ a ++ b 
    def isMix(abab: A, b: A) = 
    abab.endsWith(b.seq) && abab.startsWith(b.seq) && abab.length > 2*b.length 
} 

object MixerString extends Mixer[Char, String] { 
    val bf = implicitly[CanBuildFrom[String,Char,String]] 
    def toSeqLike(in: String): SeqLike[Char,String] = in 
} 
object MixerListInt extends Mixer[Int, List[Int]] { 
    val bf = implicitly[CanBuildFrom[List[Int],Int,List[Int]]] 
    def toSeqLike(in: List[Int]): SeqLike[Int,List[Int]] = in 
} 

Answer 1

快速的答案，你必須使用.seq像abab.endsWith(b.seq)和abab.startsWith(b.seq)。