Thread: converting 64 bit time and date

I promise this will be my last post on this topic, Ive tried that code now and im getting an "incompatible type conversion " this is driving me nuts


editted below

Code:

DATE dt;
SYSTEMTIME stime;
          FILETIME filetime;

/*******************************************************/
   
        memcpy(&filetime, createdTime, sizeof(DATE));



  FileTimeToSystemTime( &filetime, &stime ); // convert to structure

cout << stime.wMonth << "/" << stime.wDay << "/ " << stime.wYear << "     Time:"  <<           stime.wHour<<":"<< stime.wMinute << ":"<<stime.wSecond  

<<"\n";

Ive got this to compile but the date and time is not correct.....IM SURE IM CLOSE!!

> memcpy(&filetime, createdTime, sizeof(DATE));
I didn't notice this earlier, but the third argument should be sizeof(filetime) or sizeof(FILETIME):
memcpy(&filetime, createdTime, sizeof(filetime));

>im getting an "incompatible type conversion "
On which line? Just taking a quick glance I don't see it.

Also I think you can remove the DATE dt; line, unless you think you may need it in the future.

Also, if it's still not giving the expected time value, it could be because the bytes are flipped in the wrong order. I don't know how you read the timestamp from the file, but you could do it like this:

Code:

	string filename = "file.txt";	//Make this the name of your file
	FILETIME filetime;
	SYSTEMTIME stime;

	ifstream in(filename.c_str(), ios::binary);
	if (!in.is_open())

	{
		cerr << "Unable to open file: " << filename << endl;
		return 1;
	}
	in.read(reinterpret_cast<char *> &filetime, sizeof(filetime));

	FileTimeToSystemTime( &filetime, &stime ); // convert to structure

Or you can try flipping around {_ct1,_ct2,_ct3,_ct4,_ct5,_ct6,_ct7,_ct8 in the array, and see what happens.

>ive just tried both without any luck
Try printing out the sizeof(time_t) and see what value it prints. If it's 8, then maybe the timestamp is a time_t. If it's something like 4, then I don't know. I'm not familiar with Windows timestamps.

Code:

#include <ctime>
#include <iostream>
int main()
{
	std::cout << sizeof(time_t) << std::endl;

	return 0;
}

What value for year did you get cpjust? For your timestamp, I'm getting 1927 for the year. For pastitprogram's timestamp, I'm getting 1602. Which leads me to believe the original timestamp is in question ('CB 6E 70 FE B9 C8 01 00').

>After putting that in my program it printed out '4'
Right, it can't be a time_t. Are the numbers you posted the actual timestamp ('CB 6E 70 FE B9 C8 01 00')?

Originally Posted by cpjust

Yes, for the timestamp I used starting with "01" I get 1927. If I used the "CB" timestamp FileTimeToSystemTime() fails with an error code of 87 (Invalid parameter).

Ok, thanks for the info. It looks like pastitprogram got it working thanks to the code you posted.

thanks again for all the help guys, as with your example of the time which I think could work alot better than mine I have my time stamp store in the array

Code:

char createdTime[] = {_ct1,_ct2,_ct3,_ct4,_ct5,_ct6,_ct7,_ct8, '/0'};

where as you have

Code:

std::string hexTime( "01 6E 70 FE B9 C8 01 00 " );

I've been trying to insert mine as

Code:

std::string hexTime =createdTime;

It compiles but then gives me crazy output so i'm guessing this is not the right way to go about it. I was wondering if it is something to do with my declaration or theb fact my CreatedTime array is storing each byte of the time stamp in its ascii format and not in the hex characters as you have in your string?

>CreatedTime array is storing each byte of the time stamp in its ascii format and not in the hex characters as you have in your string?
I think it's actually the opposite. Your array stores 2 hex digits in one byte, where the std::string stores 2 hex digits in 2 bytes. If you want to use cpjust's code, then convert your createdTime to a std::string:

Code:

std::ostringstream converter;
int len = strlen(createdTime);
for (int i=0; i<len; i++)
{
	converter << hex << (int) createdTime[i] << " ";
}
std::string hexTime = converter.str();
std::cout << hexTime << '\n';

If you print out hexTime at this point, it should look something like what you show above. Just add the above code right below this line:
char createdTime[] = {_ct1,_ct2,_ct3,_ct4,_ct5,_ct6,_ct7,_ct8, '/0'};
Be sure to #include <cstring> for the strlen() function.