在线搜索答案会给出两个重要的帖子(Codacy's和Daniel Westheide's),两者都给出与Scala's official documentation for Try相同的答案:
上面示例中显示的Try的一个重要属性是它能够管理或链接操作,并在此过程中捕获异常。
上面引用的例子是:
import scala.io.StdIn
import scala.util.{Try, Success, Failure}
def divide: Try[Int] = {
val dividend = Try(StdIn.readLine("Enter an Int that you'd like to divide:\n").toInt)
val divisor = Try(StdIn.readLine("Enter an Int that you'd like to divide by:\n").toInt)
val problem = dividend.flatMap(x => divisor.map(y => x/y))
problem match {
case Success(v) =>
println("Result of " + dividend.get + "/"+ divisor.get +" is: " + v)
Success(v)
case Failure(e) =>
println("You must've divided by zero or entered something that's not an Int. Try again!")
println("Info from the exception: " + e.getMessage)
divide
}
}
但我可以使用传统的try
块轻松管道操作:
def divideConventional: Int = try {
val dividend = StdIn.readLine("Enter an Int that you'd like to divide:\n").toInt
val divisor = StdIn.readLine("Enter an Int that you'd like to divide by:\n").toInt
val problem = dividend / divisor
println("Result of " + dividend + "/"+ divisor +" is: " + problem)
problem
} catch {
case (e: Throwable) =>
println("You must've divided by zero or entered something that's not an Int. Try again!")
println("Info from the exception: " + e.getMessage)
divideConventional
}
(注意:divide
和divideConventional
的行为略有不同,因为后者在出现问题的第一个错误时出错,但就是这样。尝试输入“10a”作为{{1的输入看看我的意思。)
我试图看到dividend
的流水线优势,但对我而言,似乎这两种方法是相同的。我错过了什么?
答案 0 :(得分:6)
我认为你很难看到Try[T]
的构图能力,因为你在两种情况下都在本地处理异常。如果您想通过其他操作撰写divideConventional
该怎么办?
我们有一些喜欢:
def weNeedAnInt(i: Int) = i + 42
然后我们会有:
weNeedAnInt(divideConventional())
但是,让我们说你想要最大化你允许用户输入的重试次数(这通常是你在现实场景中所拥有的,你可以永远重新输入一个方法?我们必须另外用weNeedAnInt
包裹try-catch
本身的调用:
try {
weNeedAnInt(divideConventional())
} catch {
case NonFatal(e) => // Handle?
}
但是,如果我们使用divide
,并且让我们说它没有在本地处理异常并向外传播内部异常:
def yetMoreIntsNeeded(i: Int) = i + 64
val result = divide.map(weNeedAnInt).map(yetMoreIntsNeeded) match {
case Failure(e) => -1
case Success(myInt) => myInt
}
println(s"Final output was: $result")
这不简单吗?也许,我认为这有一些主观性的答案,我发现它更清洁。想象一下,我们有很长时间的这种操作,我们可以将每个Try[T]
组合到下一个,并且只在管道完成时担心问题。