Can I intercept the F# sequence generator?

Question

trying to work around a problem in outside library - is there a way to try-catch the generator itself item by item (probably not, but just to be sure...)?

let myTest() =
    let mySeq = seq { for i in -3 .. 3 -> 1 / i }
    // how to keep the line above intact, but modify the code below to try-catch-ignore the bad one?
    mySeq |> Seq.iter (fun i -> printfn "%d" i)
    ()

Answer 1

You can't.
Once the exception happens, the state of the source enumerator is screwed up. If you can't get into the source enumerator to "fix" its state, you can't make it keep producing values.

You can, however, make the whole process "stop" after the exception, but you'll have to go a level below and work with IEnumerator<T>:

let takeUntilError (sq: seq<_>) = seq {
    use enm = sq.GetEnumerator()
    let next () = try enm.MoveNext() with _ -> false
    let cur () = try Some enm.Current with _ -> None

    while next() do
      match cur() with
      | Some c -> yield c
      | None -> ()
  }

mySeq |> takeUntilError |> Seq.iter (printf "%d")