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Idiomatic way to yield "zero or one"/optional value?
f#compiler

I'm a F# newbie very much enjoying sequence expressions.

Nice and beautiful:

To yield a sequence of elements:

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yield! SomeElements(...)

To yield a single element:

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yield SomeElement(...)

But what is the best way to yield an optional element?

In my imagination, I can envision something like:

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yield? MaybeElement(...)

I know I can use:

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2
3

 match MaybeElement(...) with
 |Some(v)-> yield v
 |None -> ()

but that seems quite "noisy", is it possible to shorten this?

I could use something like this:

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3
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let ToSeq(value: option< 'a>) = 
    match value with
    | Some(v) -> [v]
    | None -> []

and then use:

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yield! ToSeq(MaybeElement(...))

but I fear that is not good from a performance perspective as it "hides" from the compiler the knowledge that the cardinality of the sequence is never more than one.

Cheers,

Jarle Stabell

.

what is wrong with 

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yield MaybeElement(...)
By on 12/20/2008 6:38 PM ()

what is wrong with
yield MaybeElement(...)

This would yield 'a option type.

How about 

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[if Option.is_some MaybeElement(...) then yield MaybeElement(...).Value];

If it is a big calculation then the others are better.

But since the size always will be 0 or 1 I can't follow why you want to move over to a sequence. This must be because of some iteration or concatination of several 0-1 list, but by keeping the match statement you also keep the redability.

By on 12/21/2008 11:10 AM ()

Moving to a sequence was only in an attempt to see if I was able to shorten the original expression.

But it's not something I'd use as it muddles the intent and probably hurts performance.

By on 12/21/2008 4:58 PM ()

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