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下最後語句行產生一個錯誤: 「類型不匹配;發現的:需要TestExpandableWithLibrary.this.library.type(與下面類型org.typeclass.Library):V」Scala的類型類和隱式的益智
這是我試圖做隱式轉換的地方。前面的行我簡單地使用typeclass函數確實工作正常。
關於如何解決它的任何想法?
package org.typeclass
///////////////////////////////////////////////////////////////////////////////
// the domain objects
case class Book(bookName: String)
case class Library(libraryName: String, books: Set[Book])
object Library {
def apply(libraryName: String, bookNames: String*): Library =
Library(libraryName, bookNames.map(Book(_)).toSet)
}
case class TreeClass(nodeName: String, children: Seq[TreeClass])
///////////////////////////////////////////////////////////////////////////////
// the typeclass definition
trait Expandable[T, V, R] {
def expandWith(template: T, values: V): R
}
object Expandable {
def expandWithF[T, V, R](template: T, values: V)(implicit ev: Expandable[T, V, R]): R =
ev.expandWith(template, values)
implicit class ExpandableItem[T, V, R](val template: T) extends AnyVal {
def expandWithM(values: V)(implicit ev: Expandable[T, V, R]): R =
ev.expandWith(template, values)
}
}
///////////////////////////////////////////////////////////////////////////////
// a typeclass implementation
object ExpandableImpls {
implicit object ExpandableTreeClass extends Expandable[TreeClass, Library, TreeClass] {
def expandWith(template: TreeClass, library: Library): TreeClass = {
val parentName = s"${template.nodeName}.${library.libraryName}"
val children = library.books.map(book => TreeClass(s"${parentName}.${book.bookName}", Seq.empty)).toSeq
TreeClass(parentName, children)
}
}
}
//@RunWith(classOf[JUnitRunner])
class TestExpandableWithLibrary /*extends FlatSpec*/ {
import Expandable._
import ExpandableImpls._
val library = Library("test", "black", "white")
val root = TreeClass("parent", Seq.empty)
val useF = expandWithF(root, library) // this works
val useM = root.expandWithM(library) // this doesn't work!
}
完美,非常感謝! – satyagraha 2014-09-10 16:49:53
不要粗暴,但也許在理想情況下,編譯器可能會給我一些關於我要去哪裏漂泊的線索。鍵入推論=龍在這裏! – satyagraha 2014-09-10 21:42:28
@satyagraha:我同意這個錯誤可能會更好。但是,如果你想要推斷出一個類型參數,那麼它應該出現在參數的非通用位置,這是一個好主意。 – 2014-09-11 13:40:00