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let SeqOf2Nodes(node1:#Node, node2:#Node) =  List.to_seq( [ (node1 :> Node); (node2 :> Node) ] )

The return type is "seq<Node>", where Node is my custom type. I am wondering if there is a way to directly make these two nodes a sequence, rather than creating a list first.

Well, a "sequence" (represented as 'seq') in F# is really just an alias for the .NET System.Collections.Generic.IEnumerable<T> interface. Since, F# lists, implement IEnumerable<T>, it's a simple type-cast:

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let SeqOf2Nodes(node1 : #Node, node2 : #Node) =
  [node1; node2] :> seq<Node>

That typecast is so common that the F# libraries provide a function to perform it:

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let seq (x : #seq<_>) = (x :> seq<_>)

So, you can just write this:

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let SeqOf2Nodes(node1 : #Node, node2 : #Node) =


  seq [node1; node2]

Nothing else is needed. If for some reason, you *really* want a lazy sequence, I suppose you could use a sequence expression:

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let SeqOf2Nodes1(node1 : #Node, node2 : #Node) =
  seq {
        yield node1
        yield node2
      }

Hmm. It's not more concised, but you can use Seq.cons to avoid creating a list:

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let SeqOf2Nodes(node1:#Node, node2:#Node) =
 Seq.empty |> Seq.cons (node2 :> Node) |> Seq.cons (node1 :> Node)