Home > SQL Server > What is the SQL Server 2008 DateTime2 Internal Structure?

What is the SQL Server 2008 DateTime2 Internal Structure?

SQL Server has a number of new date time formats, but the one I am most interested in is DateTime2. The internal format of the SQL DateTime is commonly mistaken as 2×4 byte integers, with the latter integer being milliseconds since midnight. It is in fact the number of 1/300ths of a second since midnight which is why the accuracy of the DateTime within SQL Server has historically been 3.33ms. (If you really want to see it, crack it open by converting it to a binary, adding 1 and re-converting, you add 3.33ms, not 1 ms.)

So DateTime2 must use a different format, and as a weekend exercise that had no purpose than understanding the internals I thought I’d take a look. I have not seen the information in the BoL or posted as yet, so might be of use.  I am starting with the DateTime2(7) and looking at the maximum accuracy structure. The code used to crack it open each time is basically as follows:

declare @dt datetime2(7)
set @dt = '2000/01/01 00:00:00'
declare @bin varbinary(max)
set @bin = CONVERT(varbinary(max), @dt)

To make my life easier, SQL conveniently outputs all the values as hexi-decimal numbers. The results are not what you would expect.


The date which traditionally occupied the first 4 bytes, clearly is occupying the last few bytes. So the format is not going to be obvious or simple. Interestingly the returned result is 9 bytes, but the length is quoted as 8. It is returning 8 when checked using the length, that first byte is somewhat odd to make an appearance.  It’s also suspiciously the accuracy value, and with a few tests using a change of accuracy, it show that value changes. So the first pseudo-byte is the accuracy indicator.

To start figuring out some more, let’s take the time back to the beginning point, which in this case is not 1900/01/01 but 0001/01/01 which when converted gives us:

'0001/01/01 00:00:00' => 0x070000000000000000

Start incrementing the day portion and there is an obvious pattern, the 6th byte changes.

'0001/01/02 00:00:00' => 0x070000000000010000
'0001/01/03 00:00:00' => 0x070000000000020000
'0001/01/31 00:00:00' => 0x0700000000001E0000

As you try the 2nd month, to check where the month is, the same byte alters, so it represents days, not specific date parts. Is it the number of days since the beginning of the year? No.

'0001/02/01 00:00:00' => 0x0700000000001F0000

If it was, there would be an issue since 1 byte does not represent enough values, as we can see, FF occurs on the 13th of September, and then it rolls over and puts a 1 in the 7th Byte position.

'0001/09/13 00:00:00' => 0x070000000000FF0000
'0001/09/14 00:00:00' => 0x070000000000000100
'0001/09/15 00:00:00' => 0x070000000000010100

It rolls over, then carries on as before. This immediately suggests the next test, to roll over the year, and the pattern continues.

'0001/12/31 00:00:00' => 0x0700000000006C0100  
'0001/12/31 00:00:00' => 0x0700000000006D0100

So the format is just counting, we see it in the example as hex, but it is a straight number count going on but the hex values are left-to-right. Only 2 bytes are used so far, which do not represent enough day combinations, add the third byte in by going past 180 years:

'0180/06/06 00:00:00' => 0x070000000000FFFF00  
'0180/06/07 00:00:00' => 0x070000000000000001

So the final byte is then increased, so the number of combinations becomes 16777215 – that seems a lot better and certainly going to cover the range required.

'2001/01/01 00:00:00' => 0x07000000000075250B

So that is the final 3 bytes decoded, a simple pattern – and provides the template of how the time is also stored.

'0001/01/01 00:00:00.0000000' => 0x070000000000000000
'0001/01/01 00:00:00.0000001' => 0x070100000000000000
'0001/01/01 00:00:00.0000255' => 0x07FF00000000000000
'0001/01/01 00:00:00.0065535' => 0x07FFFF000000000000
'0001/01/01 00:00:00.0065536' => 0x070000010000000000

So to check whether the format is the same,

'0001/01/01 00:00:00.9999999' => 0x077F96980000000000

Decode that again and it all matches:

select (152 * 256 * 256) + (150 * 256) + 127

When we click over into 1 second exactly, we increment the first byte by 1, so the time portion is still represented in 100ns intervals, with the normal system of each byte counting up 1 every time the previous byte rolls over. As we get to the limit of the 3 bytes, it rolls into the 4th and then the 5th.

'0001/01/01 00:00:01.0000000' => 0x078096980000000000

So the internal format of the DateTime2(7) is decoded, not difficult but it is an interesting choice – it is now a straight binary number, with the Least Significant Byte being on the Left, the Most Significant being on the right (for each section.) Within the byte however, to convert it you must still read it right-to-left.

The first 5 bytes are recording how many time units intervals have passed since midnight, and the last 3 bytes recording how many days have passed since 0001/01/01.

The time unit intervals are dictated by the accuracy of the number, 100ns for DateTime2(7), and 1 Micro second intervals for a DateTime2(6) etc.  The way in which you interpret it does not change, but the units you are multiplying the time portion by, alters based on the accuracy.

You could construct 2 dates that are identical at a binary level, but due to the field meta-data on accuracy, they do not represent the same date time.

declare @dt1 dt1 datetime2(6)
set @dt1 = '0001/01/01 00:00:00.000001'
declare @dt2 datetime2(7)
set @dt2 = '0001/01/01 00:00:00.0000001'


 And that is perhaps why on output they automatically have prefixed the binary value with the datetime accuracy, so that they are not entirely identical? I’m not sure but would be interested to find out.

  1. BEL8490
    October 13, 2009 at 9:14 am

    Thanks for this clarifying insight.

    I would suggest two minor editing corrections:

    1. The code at the beginning of the article does not have a declaration for the variable @bin1 and should be corrected as follows:
    set @bin = CONVERT(varbinary(max), @dt)

    2. At about halfway down the following example

    ‘0180/06/06 00:00:00’ => 0x070000000000FFFF00
    ‘0180/06/07 00:00:00’ => 0x070000000000FFFF00

    should be corrected as:

    ‘0180/06/06 00:00:00’ => 0x070000000000FFFF00
    ‘0180/06/07 00:00:00’ => 0x070000000000000001

    as clearly two different dates can not have the same internal presentation.

    • October 13, 2009 at 9:30 am

      Your absolutely right, I’ve ended up copying the wrong thing out of the Query window and duplicated a result that should be different. I’ve updated the post with the correction.

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