Hi Marc,
The best way of answering such questions is to check the source code.
Hackage provides
a nice way of doing that - click on 'Source' near the instance that
you are interested in:
https://hackage.haskell.org/package/time-1.6.0.1/docs/Data-Time-Format.html#...
And you'll see the implementation
```
instance ParseTime LocalTime where
buildTime l xs = LocalTime <$> (buildTime l xs) <*> (buildTime l xs)
```
That builds time from `Day` and `TimeOfDay` passing your parse string
to each of those.
Then you can check ParseTime instance of Day:
https://hackage.haskell.org/package/time-1.6.0.1/docs/src/Data.Time.Format.P...
I'm not providing it here, as it's quite big, but the main point is
that `s` is ignored so in that case
Day appear to be:
```
rest (YearMonth m:_) = let
d = safeLast 1 [x | MonthDay x <- cs]
in fromGregorianValid y m d
```
with y=m=d=1
if you continue the process for TimeOfDay you'll find that `s` is
ignored there as well, and
`midnight = TimeOfDay 0 0 0` is returned in that case.
So it appeared that LocalTime consists of the components that ignore
your parse string and return
default value instead.
I don't know if that is intended behaviour or not, but for me it makes
more sense to parse to UTCTime/POSIXTime
and then convert into LocalTime, in case if you get seconds as input.
Hope that helps.
On Thu, 6 Sep 2018 at 13:42, Marc Busqué
In GHCi
``` :m +Data.Time parseTimeM True defaultTimeLocale "%s" "1535684406" :: Maybe UTCTime -- => Just 2018-08-31 03:00:06 UTC parseTimeM True defaultTimeLocale "%s" "1535684406" :: Maybe LocalTime -- => Just 1970-01-01 00:00:00 ```
Why? ¯\(°_o)/¯
Marc Busqué http://waiting-for-dev.github.io/about/_____________________________________... Haskell-Cafe mailing list To (un)subscribe, modify options or view archives go to: http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe Only members subscribed via the mailman list are allowed to post.
-- Alexander