Question

当我提出这个想法时，我正试图在项目（Play Framework 2.4）中重构我的scala代码：

（为了提供一个最小的工作示例，我已经更改了一些类，例如，我分别使用Int和Option [Int]更改了Result和Future [Result]）

object ParFuncApply {

  trait CanBeAuthenticatedRequest[A]
  trait AuthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
  trait UnauthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]


  private def fold[T](authenticated: (AuthenticatedRequest[_]) => T)
                             (unauthenticated: (UnauthenticatedRequest[_]) => T):
  PartialFunction[CanBeAuthenticatedRequest[_], T] = {
    case ar: AuthenticatedRequest[_] => authenticated(ar)
    case ur: UnauthenticatedRequest[_] => unauthenticated(ur)
  }

  def apply(request: CanBeAuthenticatedRequest[_])
           (authenticated: (AuthenticatedRequest[_]) => Int)
           (unauthenticated: (UnauthenticatedRequest[_]) => Int): Int = {
    fold(authenticated)(unauthenticated)(request)
  }

  def async(request: CanBeAuthenticatedRequest[_])
           (authenticated: (AuthenticatedRequest[_]) => Option[Int])
           (unauthenticated: (UnauthenticatedRequest[_]) => Option[Int]): Option[Int] = {
    fold(authenticated)(unauthenticated)(request)
  }
}

上面的代码编译。

然后我说：我应该将fold [T]参数化类型限制为Int和Option [Int]，所以我补充说：

object ParFuncApply {

  trait CanBeAuthenticatedRequest[A]
  trait AuthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
  trait UnauthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]

  sealed trait Helper[T]

  object Helper {
    implicit object FutureResultHelper extends Helper[Option[Int]]
    implicit object ResultHelper extends Helper[Int]
  }

  private def fold[T: Helper](authenticated: (AuthenticatedRequest[_]) => T)
                             (unauthenticated: (UnauthenticatedRequest[_]) => T):
  PartialFunction[CanBeAuthenticatedRequest[_], T] = {
    case ar: AuthenticatedRequest[_] => authenticated(ar)
    case ur: UnauthenticatedRequest[_] => unauthenticated(ur)
  }

  def apply(request: CanBeAuthenticatedRequest[_])
           (authenticated: (AuthenticatedRequest[_]) => Int)
           (unauthenticated: (UnauthenticatedRequest[_]) => Int): Int = {
    fold(authenticated)(unauthenticated)(request)
  }

  def async(request: CanBeAuthenticatedRequest[_])
           (authenticated: (AuthenticatedRequest[_]) => Option[Int])
           (unauthenticated: (UnauthenticatedRequest[_]) => Option[Int]): Option[Int] = {
    fold(authenticated)(unauthenticated)(request)
  }
}

但是，如果我更改，此代码不再编译 fold(authenticated)(unauthenticated)(request)到fold(authenticated)(unauthenticated).apply(request)（我添加了一个显式调用apply（））它编译。为什么会这样？对类的调用（）和.apply（）应该是相同的，不是吗？

编译器似乎要求将返回类型（Int或Option [Int]）传递给PartialFunction而不是CanBeAuthenticatedRequest类型。

Answer 1

因为您在`fold [T：Helper]'中定义了一个上下文，编译器将添加另一个参数列表。换句话说，上下文绑定只是语法糖：

private def fold[T](authenticated: (AuthenticatedRequest[_]) => T)
                   (unauthenticated: (UnauthenticatedRequest[_]) => T)
                   (implicit helper: Helper[T): PartialFunction[CanBeAuthenticatedRequest[_], T]

所以当你打电话时

fold(authenticated)(unauthenticated)(request)

编译器认为request应该是一个明确指定的隐式Helper [T]。

应用带有（）的PartialFunction的Scala与.apply（）不同

1 个答案: