scala - 如何处理Reader[A, Future[B]]

Question

鉴于以下 3 个函数，

def f[A, B] : Reader[A, B] = ???
def g[A, B, C](b: B) : Reader[A, Future[C]] = ???
def h[A, C, D](c: C) : Reader[A, D] = ???

如何编写一元组合器？

这是我目前的解决方案，但我对所有这些样板并不满意

def f2[A, B] : ReaderT[Future, A, B] = kleisli {
  a => Future.successful(f.run(a))
}

def h2[A, C, D](c: C) : ReaderT[Future, A, D] = kleisli {
  a => Future.successful(h(c).run(a))
}

def g2[A, B, C](b: B) : ReaderT[Future, A, C] = kleisli {
  a => g(b).run(a)
}

def i[A,B,C,D] : ReaderT[Future, A, (B,C,D)] =
  for {
    b <- f2[A, B]
    c <- g2[A, B, C](b)
    d <- h2[A, C, D](c)
  } yield (b,c,d)

Answer 1

def i[A,B,C,D] : ReaderT[Future, A, (B,C,D)] =
  for {
    b <- f[A, B].lift[Future]
    c <- g[A, B, C](b).mapK[Future, C](identity)
    d <- h[A, C, D](c).lift[Future]
  } yield (b,c,d)

我不确定是否有更好的方法来替换 mapK[Future, C](identity) .