dollar/lambda expression layouting #16

New Issue

2017-03-15T00:03:31+01:00

lspitzner commented

2017-03-15 00:03:31 +01:00

(Migrated from github.com)

foldrDesc f z = unSwitchQueue $ \q ->
  switch (Min.foldrDesc (f . unTaggedF) z q) (Min.foldrAsc (f . unTaggedF) z q)

currently is layouted as

foldrDesc f z =
  unSwitchQueue
    $ \q ->
        switch (Min.foldrDesc (f . unTaggedF) z q)
               (Min.foldrAsc (f . unTaggedF) z q)

~~~~.hs foldrDesc f z = unSwitchQueue $ \q -> switch (Min.foldrDesc (f . unTaggedF) z q) (Min.foldrAsc (f . unTaggedF) z q) ~~~~ currently is layouted as ~~~~.hs foldrDesc f z = unSwitchQueue $ \q -> switch (Min.foldrDesc (f . unTaggedF) z q) (Min.foldrAsc (f . unTaggedF) z q) ~~~~

int-index commented

2017-03-15 00:06:59 +01:00

(Migrated from github.com)

This can also happen with operators different than ($), no?

This can also happen with operators different than `($)`, no?

lspitzner commented

2017-03-15 00:11:32 +01:00

(Migrated from github.com)

correct. i think the cause is the lambda expression anyways, because

foo = abc $ def $ do
  foo
  bar

is done properly.

correct. i think the cause is the lambda expression anyways, because ~~~~.hs foo = abc $ def $ do foo bar ~~~~ is done properly.

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Reference: hexagoxel/brittany#16