Re: [Haskell-cafe] List instance of Alternative: why (++)?

My understanding of the proposed instance was that if xs >>= f is [], then (xs >>= f) | (ys >>= f) would be ys >>= f. That may not necessarily be empty. Consider: f 42 = [42] f _ = [] xs = [1,2,3] ys = [42] On Sun, May 28, 2017 at 12:24 AM, Damian Nadales wrote:

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It's hard to look at laws, because there's apparently little agreement on the proper laws for Alternative. It looks possible that as an Applicative and Alternative, this would be fine; but the Alternative instance you propose would work in odd ways with the Monad instance. That is, if f x == [] for any x in (non-empty) xs, then something like (xs <|> ys) >>= f would yield an empty list, while (xs >>= f) <|> (ys >>= f) would not. But, this

In this case `(xs >>= f) <|> (ys >>= f)` will also be empty as far as I can see...

I'm also trying to make some sense out of this definition of alternative for lists. For `Maybe` we also have a left biased alternative, and despite this I find it quite useful...

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isn't a law or anything, you could chalk it up as counter-intuitive, but not disqualifying.

On Fri, May 5, 2017 at 11:12 PM, Theodore Lief Gannon wrote:

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Fiddling around, I found myself wanting:

coalesce :: [a] -> [a] -> [a] -- or -- :: (Foldable t) => t a -> t a -> t a coalesce a b = if null a then b else a

I expected this to be (<|>) (it is for Maybe!) but instead I find no canonical implementation of it anywhere, and what seems like a useless instance Alternative []. What's the rationale?

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