重试返回Future的函数 [英] Retry a function that returns a Future

问题描述

假设我有一个函数 foo ，该函数执行异步计算并返回 Future ：

Suppose I've got a function foo that performs asynchronous computation and returns a Future:

def foo(x: Int)(implicit ec: ExecutionContext): Future[Int] = ???

现在我想重试此计算 n 次直到计算成功。

Now I'd like to retry this computation n times until the computation succeeds.

def retryFoo(x: Int, n: Int)(implicit ec: ExecutionContext): Future[Int] = ???

我还要返回重试时抛出的所有 all 异常。因此，我定义了新的异常类 ExceptionsList 并要求 retryFoo 返回 ExceptionsList 当所有重试均失败时。

I would like also to return all exceptions thrown while retrying. So, I define new exception class ExceptionsList and require retryFoo to return ExceptionsList when all retries fail.

case class ExceptionsList(es: List[Exception]) extends Exception { ... }  

您如何编写 retryFoo 重试 foo 并返回 Future 或 foo 结果或 ExceptionsList ？

How would you write retryFoo to retry foo and return a Future with either the foo result or ExceptionsList ?

推荐答案

我可能会做这样的事情：

I'd probably do something like this:

final case class ExceptionsList(es: List[Throwable]) extends Throwable
def retry[T](n: Int, expr: => Future[T], exs: List[Throwable] = Nil)(implicit ec: ExecutionContext): Future[T] =
  Future.unit.flatMap(_ => expr).recoverWith {
    case e if n > 0 => retry(n - 1, expr, e :: exs)
    case e => Future.failed(new ExceptionsList(e :: exs))
  }

expr 是按名称调用的，因为它本身可能引发异常，而不是返回失败的Future。我将累积的例外记录在列表中，但我想那是一种味道。

expr is call-by-name since it itself could throw an exception rather than returning a failed Future. I keep the accumulated exceptions in a list, but I guess that's a taste-thing.

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