What's wrong with what you just wrote?
Also, I think you probably meant:
forever (await >>= \x -> yield (Left x) >> yield (Right x))
On 4/7/15 5:03 AM, Antonio Nikishaev wrote:
What is the sane way to do something like this?
(await >>= \x -> yield (Left x) >> yield (Right x))
>-> p1 +++ p2
Let's assume p1 or/and p2 are too complex to fit into Foldl.
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