this post was submitted on 14 Dec 2023
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https://xkcd.com/2867

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It's not just time zones and leap seconds. SI seconds on Earth are slower because of relativity, so there are time standards for space stuff (TCB, TGC) that use faster SI seconds than UTC/Unix time. T2 - T1 = [God doesn't know and the Devil isn't telling.]

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[–] ericbomb@lemmy.world 123 points 1 year ago* (last edited 1 year ago) (1 children)

We use datediff in sql and let God handle the rest.

"Oh but they're in different time zones" "Oh did you account for if one is in day light savings and other isn't" "Aren't some of these dates stored in UTC and some local?"

Are all problems I do not care about.

[–] Cosmonaut_Collin@lemmy.world 49 points 1 year ago (3 children)

This is why we should just move to a universal time zone and stop with the day light savings.

[–] nxdefiant@startrek.website 50 points 1 year ago* (last edited 1 year ago) (4 children)

We have that, it's called Unix time, and the only thing it doesn't account for is time dilation due to relativity.

it's perfect

[–] phoneymouse@lemmy.world 19 points 1 year ago (2 children)

If your system hasn’t been upgraded to 64-bit types by 2038, you’d deserve your overflow bug

[–] Appoxo@lemmy.dbzer0.com 8 points 1 year ago (1 children)

Let's just nake it 128-Bit so it's not our problem anymore.
Hell, let's make it 256-Bit because it sounds like AES256

[–] phoneymouse@lemmy.world 16 points 1 year ago* (last edited 1 year ago) (2 children)

64 bits is already enough not to overflow for 292 billion years. That’s 21 times longer than the estimated age of the universe.

[–] nybble41@programming.dev 13 points 1 year ago (2 children)

If you want one-second resolution, sure. If you want nanoseconds a 64-bit signed integer only gets you 292 years. With 128-bit integers you can get a range of over 5 billion years at zeptosecond (10^-21 second) resolution, which should be good enough for anyone. Because who doesn't need to precisely distinguish times one zeptosecond apart five billion years from now‽

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[–] Faresh@lemmy.ml 7 points 1 year ago* (last edited 1 year ago)

With a 128 bit integer you can represent 340 undecillion (or sextillion if you use the long scale notation) seconds, which is equivalent to 10 nonillion (or quintillion, long scale) years. The universe will long have have stopped being able to support life by then because stars stopped forming (enough time would have passed it could have happened a hundred quadrillion (a hundred thousand billion, long form) times over assuming we start counting from the birth of the universe).

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Give it a few more decades

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[–] marcos@lemmy.world 67 points 1 year ago (1 children)

From the wikipedia:

TCB ticks faster than clocks on the surface of the Earth by 1.550505 × 10−8 (about 490 milliseconds per year)

It's amazing that this level of detail is relevant to anything.

[–] elvith@feddit.de 54 points 1 year ago (5 children)
[–] hakunawazo@lemmy.world 13 points 1 year ago (2 children)

Thank you, but I gave up halfway through the list.

[–] kurwa@lemmy.world 8 points 1 year ago (1 children)

I got to "The day before Saturday is always Friday" and I was like waaaa?

[–] sukhmel@programming.dev 8 points 1 year ago (1 children)

I thought it is about when Julian calendar was dropped in favour of Gregorian, but that's not it:

Thursday 4 October 1582 was followed by Friday 15 October 1582

[–] elvith@feddit.de 9 points 1 year ago* (last edited 1 year ago) (1 children)

Also some of the islands around the International Date Line did switch their stance on which side of the Date Line they are. So... they might have had a day twice or lost a whole day in the process. And maybe, they didn't change sides only once...

E.g. see here https://youtu.be/cpKuBlvef6A

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[–] whoisearth@lemmy.ca 5 points 1 year ago

Epoch is your friend, or use UTC. At least that's my layman reasoning. I have no challenges working with DateTime except when I don't know the underlying conditions applied from the source code.

[–] randy@lemmy.ca 10 points 1 year ago (4 children)

I really wish that list would include some explanations about why each line is a falsehood, and what's actually true. Particularly the line:

The software will never run on a space ship that is orbiting a black hole.

If the author has proof that some software will run on a space ship that is orbiting a black hole, I'd be really interested in seeing it.

[–] nybble41@programming.dev 12 points 1 year ago

Technically isn't the Earth itself a sort of space ship which is orbiting (...a star which is orbiting...) the black hole at the center of the Milky Way galaxy? Not really close enough for time dilation to be a factor, but still.

[–] elvith@feddit.de 7 points 1 year ago

All links to the original article are dead and even archive.org doesn't have a capture either. I guess the argument is along the lines of "it might not be relevant, when you're scripting away some tasks for your small personal projects, but when you're working on a widely used library or tool - one day, it might end up on a space vessel to explore whatever."

E.g. my personal backup script? Unlikely. The Linux kernel? Somewhat plausible.

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[–] sukhmel@programming.dev 10 points 1 year ago (1 children)

This one is good (or evil, depends on how you see it):

Human-readable dates can be specified in universally understood formats such as 05/07/11.

[–] elvith@feddit.de 9 points 1 year ago

That one's really good.

Which one is it?

  • July 5th 2011
  • May 7th 2011
  • July 11th 2005
  • November 7th 2005

And is it 2011/2005 or rather 1911/1905, 1811/1805,...?

[–] Kethal@lemmy.world 8 points 1 year ago (1 children)

Does anyone know what is untrue about "Unix time is the number of seconds since Jan 1st 1970."?

[–] icydefiance@lemm.ee 27 points 1 year ago* (last edited 1 year ago) (1 children)

When a leap second happens, unix time decreases by one second. See the section about leap seconds here: https://en.m.wikipedia.org/wiki/Unix_time

As a side effect, this means some unix timestamps are ambiguous, because the timestamps at the beginning and the end of a leap second are the same.

[–] nybble41@programming.dev 4 points 1 year ago

It might be more accurate to say that Unix time is the number of days since Jan 1st, 1970, scaled by 24×60×60. Though it gets a bit odd around the actual leap second since they aren't spread over the whole day. (In some ways that would be a more reasonable way to handle it; rather than repeating a second at midnight, just make all the seconds slightly longer that day.)

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[–] chuck@lemmy.ca 49 points 1 year ago (3 children)

Ah I've gotten to the point where I have to define what "frame" and epoch each time base is in before I'll touch the representation of time( Unix,Gregorian, etc) .To be honest I'm probably just scratching the surface of time problem.

Hell probably the reason we haven't seen time travellers is we suck at tracking time and you probably need to accurately know your time and place to a very good precision to travel to a given point and we can't say where and when that is with enough accuracy to facilitate where to land. And people don't want to land in the earth's surface or 10000 km away from a stable orbit. Maybe some writer can build that out for a time travel book or to discount it for some reason lol

[–] kurwa@lemmy.world 19 points 1 year ago (1 children)

I recall a short story like that where someone died because they time traveled, but didn't account for position.

[–] niktemadur@lemmy.world 5 points 1 year ago* (last edited 1 year ago)

Then there's continental drift, which as Indiana Jones reminded us this past summer, Archimedes didn't know about when he built his time machine.

Pet peeve: brushing aside the time travel fantasy element, there is not a single shred of evidence of any type of connection between Archimedes and the Antikythera Mechanism.

As if the only person clever enough in Ancient Greece was that one famous dude from Syracuse.
Ionians: "Are we a joke to you?"

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[–] reverendsteveii@lemm.ee 35 points 1 year ago* (last edited 1 year ago) (12 children)

I just spent two days debugging a reporting endpoint that takes in two MM-YYYY parameters and tries to pull info between the first day of the month for param1 and the last day of the month for param2 and ended up having to set my date boundaries as

LocalDate startDate = new LocalDate(1, param1.getMonth(), param2.getYear()); //pretty straightforward, right?

//bump month by one, account for rollover, set endDate to the first of that month, then subtract one day

int endMonth = param2.month == 12 ? param2.month + 1 : 1;

LocalDate endDate = new LocalDate(1, endMonth, param2.year).minusDays(1);

This is extraordinarily simply for humans to understand intuitively, but to code it requires accounting for a bunch of backward edge/corner case garbage. The answer, of course, is to train humans to think in Unix epoch time.

[–] hikaru755@feddit.de 13 points 1 year ago (1 children)

Using YearMonth.atEndOfMonth would have been the easier choice there, I think

[–] reverendsteveii@lemm.ee 15 points 1 year ago (1 children)

holy shit, yeah it would have. tyvm, I'll be putting in a PR first thing monday!

[–] Jarix@lemmy.world 5 points 1 year ago (2 children)

Would you mind trying to explain (ELI5 style) what you did before and why you are excited for this new method for those of us who dont understand code?

[–] reverendsteveii@lemm.ee 8 points 1 year ago (6 children)

it does in a way that's been reviewed, vetted and tested by a lot of people the thing that I'm trying to do with code that's only ever been seen by me and one other guy and has been tested to this best of my ability, which i hope is quite good but one person can easily miss edge cases and weird coincidences.

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[–] drislands@lemmy.world 5 points 1 year ago* (last edited 1 year ago) (2 children)

To break it down a bit further, the code that was provided is specifically trying to get the last day of a month, which I'll call Month X since it will vary. The code is doing these things, in this order:

  1. Get the month after Month X
  2. If Month X is 12 (aka December) get Month 1 instead (aka January)
  3. Get the Date that is day 1 of the Month from step 2
  4. Get the Date that is one day before the Date from step 3

All this to get the last day of the month from Month X. The reason they did it this way is so they didn't have to say "Is this February? Then get day 28. Is this January/March/etc? then get day 31." and so on.

The code that the other user provided will instead get the last day of Month X without having to do all those steps. It's doing something in the background to get the same data, but the coder doesn't have to worry about exactly how because they can trust it will work as expected.

It ultimately boils down to the user carving out a round piece of wood, fitting it on an axle and bolting it on, then to find someone already has cheap wheels for sale that are more stable than what they just made.

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[–] drislands@lemmy.world 8 points 1 year ago (1 children)

In the example you gave, wouldn't the year be off by one when param2.month is 12?

[–] reverendsteveii@lemm.ee 9 points 1 year ago (1 children)

I was transcribing it from memory and that exact problem cost me like two hours when I was writing it the first time. Well spotted, now write me a unit test for that case.

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[–] BeautifulMind@lemmy.world 29 points 1 year ago* (last edited 1 year ago) (1 children)

LOL whenever I have to work with DateTime systems that try to account for every possibility (and fail trying) I am reminded that in some disciplines, it's acceptable to simplify drastically in order to do 'close enough' work.

I mean, if spherical cows are a thing because that makes the math of theoretical physics doable, why not relativity-free or just frame-constant date-time measures that are willing to ignore exotic edge cases like non-spherical livestock?

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[–] MxM111@kbin.social 25 points 1 year ago* (last edited 1 year ago) (2 children)

Not even going to general relativity, as this comics suggests to experience time slowing down due to gravity, the events are not just at time, but also at particular location in space relative particular inertial system. Not specifying it, and not specifying the inertial system for which the final answer is needed makes it impossible to calculate even in special relativity, without effects of gravity.

[–] steventhedev@lemmy.world 27 points 1 year ago* (last edited 1 year ago) (1 children)

Here's the shortlist of horrors I've had to deal with in my career:

  • Mixed US/ROW short date formats - DD/MM/YY, MM/DD/YY
  • mixed timezones in the same column
  • the wrong timezone (marked as PDT but actually UTC, or sometimes the other way around)
  • clock drift
  • timezones again...because timezones suck
  • historical timezones
  • NTP configurations

Things I've read about but haven't needed to deal with personally:

  • leap seconds
  • clock slew vs skip
  • hardware clocks
  • PTP

The one thing I really don't care about is relativity

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[–] JohnDClay@sh.itjust.works 20 points 1 year ago (2 children)

Relevant Tom Scott video the answer depends on location, times, country, religion, and a bunch of other factors.

[–] threelonmusketeers@sh.itjust.works 4 points 1 year ago* (last edited 1 year ago) (1 children)

I wonder what fraction of xkcd readers have already seen this video... I imagine it is quite high. I know I've watched it multiple times. It's a good one.

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[–] randomaccount43543@lemmy.world 16 points 1 year ago
[–] Limonene@lemmy.world 12 points 1 year ago (2 children)

C++ user with operator overloading: "T2 minus T1."

Let someone else implement the class. There's probably a library for it.

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