撰写尝试更好

Question

我希望有一些实用程序可以使用和清理资源,安全和不安全,并在使用后清理资源,有点相当于try/finally,即使操作抛出异常也允许清理.

我有

def withResource[R, U](create: => R, cleanup: R => Unit)(op: R => U): U = {
  val r = create
  val res = op(r)
  cleanup(r)
  res
}

def tryWithResource[R, U](create: => R, cleanup: R => Unit)(op: R => U): Try[U] = {
  val tried: (R => Try[U]) = (r: R) => Try.apply(op(r))
  withResource(create, cleanup)(tried)
}

但我不喜欢

val tried: (R => Try[U]) = (r: R) => Try.apply(op(r))

看来我缺少一些明显的构图功能,但我看不到哪里.我试过了

val tried: (R => Try[U]) = (Try.apply _).compose(op)

但是这样做没有成功

type mismatch;
[error]  found   : R => U
[error]  required: R => => Nothing
[error]     val tried: (R => Try[U]) = (Try.apply _).compose(op)

我错过了什么/做错了什么？

Answer 1

你可以使用 map

val tried: (R => Try[U]) = Try.apply(_).map(op)

http://www.scala-lang.org/files/archive/nightly/docs/library/index.html#scala.util.Try

Answer 2

您可以使用类型归属将要传递的参数限制Try.apply为U:

val tried = (Try.apply(_: U)) compose op