F#: Compose empty sequence without using Seq.empty

Question

I write my own reimplementation of LINQ using F# (thanks to Jon Skeet for inspiration).

I use a trick to generate empty sequence:

let empty<'b> =
        seq {
            for n = 0 to -1 do
                yield Unchecked.defaultof<'b>
        }

printfn "%A" empty<int> // -> seq []

Is there any idiomatic approach to do this?

(Seq.empty is not useful, I'm just reimplementing it)

why is Seq.empty not usefull while the seq builder is? Anyway you can always use an object-expression that returns an IEnumerable<'b> which returns empty IEnumerator<'b> s — Random Dev
– Random Dev, Commented Oct 19, 2015 at 13:33
also [] :> 'a seq or [||] :> 'a seq or anything similiar ;) — Random Dev
– Random Dev, Commented Oct 19, 2015 at 13:34

Tomas Petricek · Accepted Answer · 2015-10-19 13:39:25Z

9

The simplest implementation using sequence expressions I can think of is:

let empty() = seq { do () }

Or if you want a generic value rather than a function:

let empty<'T> : seq<'T> = seq { do () }

One would want to write just seq { } for a sequence expression that does not produce any values, but that's not syntactically valid and so we need to do something inside the sequence expression. Using do () is just a way to tell the compiler that this is a syntactically valid sequence expression that does not do anything (and does not produce any values) when evaluated.

answered Oct 19, 2015 at 13:39

Tomas Petricek

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