鉴于以下3个功能,
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monadic组合器?
这是我目前的解决方案,但我并不满意所有这些样板
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)
答案 0 :(得分:2)
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)
。