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simplify

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and a bit faster using Eq this way
This commit is contained in:
Joey Hess 2019-05-01 15:34:07 -04:00
parent 63ed61ae3d
commit ec697721e4
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1 changed files with 7 additions and 13 deletions
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20
Annex/Import.hs
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@ -188,8 +188,9 @@ buildImportCommit' importcommitconfig mtrackingcommit imported@(History ti _) =
importedcommit <- case getRemoteTrackingBranchImportHistory h of
Nothing ->
mkcommits imported
Just oldimported ->
mknewcommits oldimported imported
Just oldimported@(History oldhc _) -> do
let oldimportedtrees = mapHistory historyCommitTree oldimported
mknewcommits oldhc oldimportedtrees imported
Just <$> makeRemoteTrackingBranchMergeCommit'
trackingcommit importedcommit ti
where
@ -207,19 +208,12 @@ buildImportCommit' importcommitconfig mtrackingcommit imported@(History ti _) =
parents <- mapM mkcommits (S.toList hs)
mkcommit parents importedtree
-- Reuse the commits from the oldimported History when possible.
mknewcommits old@(History oldhc _) new@(History importedtree hs)
| sameasold old new = return $ historyCommit oldhc
-- Reuse the commits from the old imported History when possible.
mknewcommits oldhc old new@(History importedtree hs)
| old == new = return $ historyCommit oldhc
| otherwise = do
parents <- mapM (mknewcommits old) (S.toList hs)
parents <- mapM (mknewcommits oldhc old) (S.toList hs)
mkcommit parents importedtree
-- Are the trees in the old History the same as the newly imported
-- trees, all the way down?
sameasold (History oldhc olds) (History importedtree hs)
| historyCommitTree oldhc /= importedtree = False
| otherwise = all (sameasold' olds) (S.toList hs)
sameasold' olds h = any (\old -> sameasold old h) (S.toList olds)
{- Builds a history of git trees reflecting the ImportableContents.
-