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Can anyone help crack this time format?

开发者 https://www.devze.com 2023-02-19 12:02 出处:网络
I have the following time samples. The time is stored in a 32-bit representation which I will depict in HEX along with the corresponding timestamp.

I have the following time samples. The time is stored in a 32-bit representation which I will depict in HEX along with the corresponding timestamp.

e2 51 14 68 = 2011-03-23 11:56:33.684237 (UTC-4)
e2 51 19 6f = 2011-03-23 11:56:33.812511 (UTC-4)
e2 51 a0 42 = 2011-03-23 11:56:37.542177 (UTC-4)
e7 25 49 4f = 2011-03-23 14:11:35.261131 (UTC-4)
e7 2e 71 0a = 2011-03-23 14:12:35.257552 (UTC-4)

Any help in correlating these binary values with the time to determine format would be greatly appreciated. Please note that the clock source is different for the binary value and the timestamp, so there coul开发者_运维百科d be a small offset or slight variation.


Using the differences:

~0.2 seconds = 1287
~3.7 seconds = 34515
~8098 seconds = 80980237
~60 seconds = 600000

so it's just a certain number of tenths of milliseconds stored as an unsigned integer.

2011-03-23 11:56:33.684237 is around 3796964456 tenths of a millisecond, which is 379696.4456 seconds or ~4.395 days. So the origin (time 0) is 2011-03-19 02:28:17.

So in summary: The number of tenths of milliseconds since about 2011-03-19 02:28:17.

All in all it won't last long.. it seems like a short-term counter which doesn't keep track of years or months so much. Its total range is ~5 days.


I subtracted E251 1468 from E72E 710A, converted it to decimal, and got 81,616,034.

I subtracted 11:56:33.684237 from 14:12:35.257552, converting to seconds, and got 8161.573315.

I'm guessing that the format is a tenth of a millisecond counter.

E251 1468 converted to decimal is 3,796,964,456. Converting to time units, I get 105 hours, 28 minutes, and 16.4456 seconds.

I can't tell what the starting point of the count is, unless there's more than a slight variation.

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