Question

我希望有一些实用程序可以安全和不安全地使用和清理资源，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))

似乎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)

但是没有使用

进行类型检查

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

您可以使用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

您可以使用类型归档将传递给Try.apply的参数限制为U：

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

撰写尝试更好

2 个答案: