Can I intercept the F # sequence generator?

Question

trying to get around the problem in an external library - is there a way to try to catch the generator element itself by element (maybe not, but just to make 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 not. As soon as an exception occurs, the state of the source enumerator is confused. If you cannot enter the original counter to “lock in” its state, you cannot force it to continue to create values.

You can, however, make the whole process “stopped” after an exception, but you need to go below the level 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")