撰写更好 [英] compose Try nicer

问题描述

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，这在某种程度上等同于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 =创建
 val res = op（r）
 cleanup（r ）
 res 
} 
 
 def tryWithResource [R，U]（create：=> R，cleanup：R => Unit）（op：R => U ）：Try [U] = {
 val尝试过：（R => Try [U]）=（r：R）=> Try.apply（op（r））
 withResource（create，cleanup）（tried）
} 
  

但是我不喜欢

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

似乎我缺少一些明显的合成功能，但是我看不到哪里。我尝试了

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

但使用

<$ p进行类型检查失败$ p> 类型不匹配；发现
[错误]：R => U
[必需的] [错误]：R => =>没有尝试
[错误] val：（R => Try [U]）=（Try.apply _）。compose（op）


我缺少什么/做错什么了？

解决方案

您可以使用一种类型限制将您传递给 Try.apply 的参数限制为 U 的参数：

  val尝试=（Try.apply（_：U））撰写op 
  

I want to have some utilities to use and cleanup a resource, safely and unsafely, and clean up the resource after use, somewhat equivalent to a try/finally, allowing for cleanup even if the operation throws an exception.

I have

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)
}

but I don't like

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

It seems i'm missing some obvious composition function, but I can't see where. I tried

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

but that fails typecheck with

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

What am I missing/doing wrong?

解决方案

You can use a type ascription to limit the parameter you pass to Try.apply to U :

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

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