How to convert 8 bytes hex to DateTime

Question

I'm attempting to parse a file that has a time stamp that i believe is 8 bytes long and looks like so in hex

00 00 00 00 DE A4 4F 4F

I do not receive the correct date/time when i parse this as an Int64. However if i skip the first 4 bytes and do something like so i get the correct datetime.

TimeSpan span = TimeSpan.FromTicks(BitConverter.ToInt32(bytes.Skip(index).Take(8).ToArray(),4) * TimeSpan.TicksPerSecond);
DateTime t = new DateTime(1970, 1, 1).Add(span);
StartTime = TimeZone.CurrentTimeZone.ToLocalTime(t);

However I'm not certain the next files i get to parse are going to have leading 00's for the first 4 bytes. If i parse this as a ToInt64 i throw an outOfRange exception. What is the proper way to parse this?

Please don't prefix your titles with "C#" and such. That's what the tags are for. — John Saunders, Mar 06 '12 at 21:01

Jon Skeet · Accepted Answer · 2012-03-06T23:14:23.973

5

Are you sure the values aren't actually

DE A4 4F 4F 00 00 00 00

... for example, due to you miscounting the position in the file?

That would make more sense, and represent a little-endian 64-bit integer, which you'd convert like so:

static readonly DateTime Epoch = new DateTime(1970, 1, 1, 0, 0, 0,
                                              DateTimeKind.Utc);

...

long seconds = BitConverter.ToInt64(data, index);
DateTime date = Epoch + TimeSpan.FromSeconds(seconds);

(Note how much simpler this conversion code is, by the way - no messing around with LINQ, and no mulitplying the value by TicksPerSecond.)

EDIT: If it's really two little-endian 32-bit integers, arranged in big-endian order to create a 64-bit integer, then you could use:

ulong x = BitConverter.ToUInt32(data, index);
ulong y = BitConverter.ToUInt32(data, index + 4);
ulong combined = (x << 32) | y;
long seconds = unchecked ((long) combined);

edited Mar 06 '12 at 23:14

answered Mar 06 '12 at 21:11

Jon Skeet

1,421,763
867
9,128
9,194

Looking at the file in a hex editor it reads 00 00 00 00 DE A4 4F 4F. I had originally tried something similar to what you have above sir, however i was getting an argument out of range exception. – poco Mar 06 '12 at 23:12
@poco: Is the file really only 8 bytes? It seems very odd. What's creating the file? – Jon Skeet Mar 06 '12 at 23:12
The file is actually fairly long sir, however i know what the bytes correspond to before and after these 8 bytes. It is possible that the first 4 bytes are just padding, however i'm pretty confident that the 8 bytes are the DateTime. You're edit makes total since and I'm kicking myself right now as I should have come up with that hours ago. Thank you very much for your help sir. – poco Mar 07 '12 at 03:17

score 0 · Answer 2 · edited May 23 '17 at 12:22

0

Based on how you are getting a valid DateTime out of the data, it looks very much like a POSIX time_t format: number of seconds since 1/1/1970, stored as a 64-bit integer value. There are several related questions on how to convert once of these to a DateTime. See if this helps you:

How to convert a Unix timestamp to DateTime and vice versa?

edited May 23 '17 at 12:22

Community

1
1

answered Mar 06 '12 at 21:06

Michael Edenfield

28,070
4
86
117

score -1 · Answer 3 · answered Mar 06 '12 at 21:05

What is the proper way to parse this?

Parsing this "properly" requires knowledge of how the binary data is stored, and what it represents exactly.

In your case, it sounds like it must be parsed as an Int32. You will need to determine the proper padding in all cases. There is no way to "know" how to "properly" parse a binary stream without knowing, in advance, the structure of the data in the stream.

As a separate issue, this statement:

BitConverter.ToInt32(bytes.Skip(index).Take(8).ToArray(),4)

Could be written more efficiently as:

BitConverter.ToInt32(bytes, index + 4)

Using LINQ in this case, just makes the operation very inefficient, since ToInt32 already provides an offset.

How to convert 8 bytes hex to DateTime

3 Answers3