hex convert to datetime

Question

Need some help to determine how is this date/time being encoded Here are some examples (known date only):

• 2B5F0200 -> 31/10/2021
• 2B9F0200 -> 31/12/2021
• 2C3F0200 -> 31/01/2022

I don't understand how this datetime format works.

Answer 1

First, since the last two bytes are the same in all cases, let's focus on the first two bytes. Look at them in binary:

0x2B5F: 0b_0010_1011_0101_1111
0x2B9F: 0b_0010_1011_1001_1111
0x2C3F: 0b_0010_1100_0011_1111

Next, consider the binary representation for the numbers in the dates. In some date formats, months are 0-based (January is 0), in others they're 1-based, so include both.

21: 0b_1_0101
22: 0b_1_0110
10: 0b_1010, 0b_1001
12: 0b_1010, 0b_1001
01: 0b_0001, 0b_0000
31: 0b_1_1111

By inspection, each of these binary numerals appears in the appropriate date. 31 is the last 5 bits. The next 4 bits are 10, 12 and 1. 21 and 22 show up in the first 7 bits (for 100 years, you'll need at least 7 bits).

                21   10    31
0x2B5F: 0b_0010101_1010_11111
                21   12    31
0x2B9F: 0b_0010101_1100_11111
                22    1    31
0x2C3F: 0b_0010110_0001_11111

The format is thus a packed bit-field:

 0                   1            
 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5
+-------------+-------+---------+
| year (YY)   | month | day     |
+-------------+-------+---------+

Or, as bit masks:

year:  0xFE00
month: 0x01E0
day:   0x001F