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 <marc@lamarciana.com> wrote:
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