如何解决Scala中的“类型不匹配”错误？

Question

我是scala的新手，正在尝试解决从使用SendGridEmailParams切换到SendGridHandlebarEmailParams时遇到的“类型不匹配”错误。这是我得到的错误：

Type mismatch, expected: ReferenceNodeMessage.GetCategoriesResponse => Future[NotInferedS], actual: ReferenceNodeMessage.GetCategoriesResponse => Any

这是我从SendGridEmailParams切换到SendGridHandlebarEmailParams的地方：

  private lazy val redemptionSendGridEmailParams = SendGridHandlebarEmailParams(
    templateName = redemptionEmailTemplateName,
    subject = "Verify ServiceProviderName completed your Category service"
  )
  private lazy val requestAQuoteSendGridEmailParams = SendGridHandlebarEmailParams(
    templateName = requestAQuoteEmailTemplateName,
    subject = "Verify ServiceProviderName completed your Category project"
  )

有人知道我为什么收到此错误以及如何解决该错误吗？