how to print the index of failure in catch expression in scala? - scala

Question

I have a code looking like this:
import scala.util.{Try, Success, Failure}
Try(
for (i <- 1 to 1000) {
doSomething(df(i))
}
) match {
case Success(t) => println(s"success")
case Failure(t) => println(s"failure")
}
i want to print the index of the failed input. how to print the index i in catch expression?

Answer 1

You can do this instead using Cats:
import scala.util.Try
import cats.implicits._
(1 to 1000).traverse(i => Try(doSomething(df(i))).toEither.left.map(ex => (ex, i))) match {
case Right(_) => println("success")
case Left((ex, i)) => println(s"failure: ${ex.getMessage} on idx: ${i}")
}
If you do not want to use Cats, you can just:
val attempts = for {
i <- Stream.range(start = 1, end = 1000) // Thanks to Bogdan for the idea of using a Stream.
} yield Try(doSomething(df(i))).toEither.left.map(ex => (ex, i))
attempts.collectFirst { case Left((ex, i)) => ex -> i } match {
case None => println("success")
case Some((ex, i)) => println(s"failure: ${ex.getMessage} on idx: ${i}")
}

Answer 2

You should definitely follow Luis's answer, but to address your comment, you could also catch IllegalArgumentException and re-throw it with the added index to the message, perhaps something like so:
Try(
for (i <- 1 to 1000) {
try doSomething(i) catch { case e: IllegalArgumentException => throw new IllegalArgumentException(s"Failed with index $i", e)}
}
) match {
case Success(t) => println(s"success")
case Failure(t) => println(s"failure", t)
}
However this seems hideous, and I do not advise it.

Answer 3

IMO the question hints that the code is lying.
you could write the code differently:
import scala.util.{Try, Success, Failure}
for (i <- 1 to 1000) {
Try(
doSomething(df(i))
) match {
case Failure(t) => println(s"failure on $i")
case _ =>
}
}
But you don't want to. Why not? Because you want to stop the iteration after the first failure. But you're using loop from 1 to 1000. You don't really intend to do the whole 1000 iterations. You're using an exception to break a for loop.
I would rewrite this code to make it clear that i don't intend to iterate the entire range explicitly.
You could, for example, use find instead of for, to find the index that causes a failure to happen. if None was found -> everything was successful.
so something similar to (untested):
(1 to 1000).indexWhere{index=>Try{doSomething(index)}.isFailure
i'm not sure if it's find or indexWhere in scala but you get the idea.
if you would like to obtain the exception as well, not just the index you could use views (https://docs.scala-lang.org/overviews/collections/views.html) to change your sequence to a lazily evaluated one, map the list to a tuple of form (index, Try) (without iterating the entire collection, due to lazyness of .view result), and then collectFirst where second element of tuple is Failure.
so something like (untested):
(1 to 1000).view.map{index => (index, doSomething(index)}.collectFirst{case (i,Failure(e)) => println(s"error was $e at index $i")}
alternatively you could write a very very small recursion to iterate the index sequence (also untested)
def findException(indexes: Seq[Int]): Option[(Int, Exception)] = indexes match {
case Nil => None
case index+:remaining =>
Try(doSomething(i)) match {
case Success(_) => findException(remaining)
case Failure(e) => Option((index,e))
}
findException(1 to 1000).map(println)
one question is how did you determine 1 to 1000?
this question would look differently if you had a collection of elements to verify, and not a range. in that case you would probably just use foldLeft.