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Home/ Questions/Q 7553913
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T11:11:37+00:00 2026-05-30T11:11:37+00:00

From a class which does not implement Enumerator I can now create one (thks

  • 0

From a class which does not implement Enumerator I can now create one (thks Daniel)

   type Bloomberglp.Blpapi.Element with
     member this.GetEnumerator() = 
       (seq { for i in 0 .. this.NumElements - 1 -> this.GetElement(i) }).GetEnumerator()

I am looking to create an IEnumerable wrapper from it

The following works, but is there a better way?

(for instance, a way to not have to specify IEnumerable interface whose implementation can derives from IEnumerable)

 member this.ToEnumerableElements():IEnumerable<Element> = {
     new IEnumerable<Element> with 
        member anon.GetEnumerator() :IEnumerator<Element> = this.GetEnumerator()
        member anon.GetEnumerator() :IEnumerator  = this.GetEnumerator() :> IEnumerator
 }
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  1. Editorial Team
    Editorial Team
    2026-05-30T11:11:39+00:00Added an answer on May 30, 2026 at 11:11 am

    If you want a ToEnumerable method you shouldn’t create a GetEnumerator method too. Generally, calling GetEnumerator directly is a code smell anyway.

    type Bloomberglp.Blpapi.Element with
      member this.ToEnumerable() = Seq.init this.NumElements this.GetElement
    

    With this method in place, you can use the Seq module for most operations and should never have to call GetEnumerator directly.

    For example:

    elmt.ToEnumerable() |> Seq.iter (printfn "%O")
    
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