How to ignore exception?

Question

Is there more elegant way to write:

try {
  ... // Some throwing code
  return first
} catch {
  case e:ExceptionType => {} // No code to execute. Ignore error.
}
return second

Answer 1

In Scala all exceptions are not checked, so if you don't want, you may just skip handling them (and thus exception will be escalated to a higher level). Silently ignoring an exception the way you want to do is generally a bad practice. However, your code can be shortened to:

try {
  ... // Some throwing code
} catch {
  case e:ExceptionType => 
}

Answer 2

scala.util.control.Exception.ignoring(classOf[ExceptionType]) {
  ... // Some throwing code
}

Answer 3

@Daniel has already provided the canonical method to use to do this. Look through the other methods in scala.util.control.Exception--they are quite helpful and generic!

If you need to get a return value out of the try block, use failing instead of ignoring (but be aware that the result is an Any, i.e. not typesafe).

You can also write your own exception-catcher, which will be a little slow for heavy-duty work but otherwise nice to use:

class DefaultOn[E <: Exception] {
  def apply[A](default: => A)(f: => A)(implicit m: Manifest[E]) = {
    try { f } catch { case x if (m.erasure.isInstance(x)) => default }
  }
}
object DefaultOn { def apply[E <: Exception] = new DefaultTo[E] }

scala> DefaultOn[NumberFormatException](0) { "Hi".toInt }
res0: Int = 0

Or if you like options:

class TryOption[E <: Exception] {
  def apply[A](f: => A)(implicit m: Manifest[E]) = {
    try { Some(f) } catch { case x if (m.erasure.isInstance(x)) => None }
  }
}
object TryOption { def apply[E <: Exception] = new TryOption[E] }

scala> TryOption[NumberFormatException] { "Hi".toInt }
res1: Option[Int] = None

Or you can be inspired by this plus the library routines and create your own methods to ignore multiple different exceptions and preserve types on the return value.

Answer 4

Exception? what exception?