Adding that case will require one to evaluate the second argument to check
it's empty before allowing one to examine the result.
Consider `x ++
some_list_that_takes_a_long_time_to_produce_its_first_element`.
In this case your proposal will not be an optimisation.
On Sun, Mar 4, 2018 at 8:32 PM, Ben Franksen
I found that in base [1] we have
(++) [] ys = ys (++) (x:xs) ys = x : xs ++ ys
I had expected there to be a clause
(++) xs [] = xs
at the top, which would avoid the list traversal in this common case.
Is this somehow magically taken care of by the
{-# RULES "++" [~1] forall xs ys. xs ++ ys = augment (\c n -> foldr c n xs) ys #-}
? No, inlining @augment g xs = g (:) xs@ gives me
(\c n -> foldr c n xs) (:) ys
= foldr (:) ys xs
which still traverses xs even if ys=[].
Any thoughts?
[1] http://hackage.haskell.org/package/base-4.10.1.0/docs/ src/GHC.Base.html#%2B%2B
Cheers Ben
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