5 Feb
2018
5 Feb
'18
4:09 a.m.
No, but it can still be written quite easily: unionsWith f = foldl' (unionWith f) M.empty Is that really hard enough to merit the API clutter? Maybe, but I doubt it. On Sun, Feb 4, 2018 at 2:47 PM, Johannes Waldmann <johannes.waldmann@htwk-leipzig.de> wrote:
2. Deprecate and remove.
Removal is good. (Less code, less errors ...)
But the following cannot be obtained via Foldable.fold ?
unionsWith :: Ord k => (a -> a -> a) -> [Map k a] -> Map k a
- J.W.