Add Zone property to ZonedClock that returns DateTimeZone of the clock

[rodion-m]

Hi there. It came as a surprise to me that ZonedClock don't have its DateTimeZone public property.
Sometimes we need to get know the timezone of given clock, please consider to add this property:

public DateTimeZone Zone => zone;

[jskeet]

I'll consider it, but could you give me more information about the use case? (That will help me work out whether we also need to expose the clock and the calendar as well.)

As a workaround for the moment, you could always call GetCurrentZonedDateTime() and check the time zone of that.

[rodion-m]

Actually I wanted to use as a part of my extension ToStringRelatively, that returns relatively string of the given interval (for instance Today 14:00 - 15:00), so it requires clock to get current time and timezone to convert Instants to Zoned/Local date time, since I use ZonedClock I had to duplicate time zone, because ZonedClock doesn't have this property.

[jskeet]

So does the method actually ask the ZonedClock for the current instant in its time zone? If you're not calling any of the time-zone-specific methods from ZonedClock, I think that taking an IClock and a DateTimeZone separately would be clearer. (It's like if you want a date and a time, but you're not actually referring to a specific LocalDateTime - I wouldn't recommend using LocalDateTime in that scenario just because it happens to combine the two.)

[jskeet]

Any more information on this? I'd really like to understand scenarios where accepting a ZonedClock and then accessing the Zone is really the best approach.

[rodion-m]

Hi John! I'm sorry for delay. Actually I solved this issue creating my own class LocalInterval (consisted of two LocalDateTimes) from two Instants like this: LocalInterval.FromInstants(instantStart, instantEnd, tz). So, it solved my use case. If I met similar issue again, I'll post it here. For now it solved. Anyway I think it's worth to open Zone field to ZonedClock, I assume it's logical and it may make life easier to someone.

[rodion-m]

Thank you!