CODEWITHSUNDEEP
c++datedatetimeboosttime
user1742892
user1742892

Reputation: 21

(Detailed) how to calculate time duration from epoch 1900 to now use boost?

As title, how to calculate time duration from epoch 1900 to now use boost?

Edit: Sorry about too short question at previous. I will describe my question again. I have problem about save birthday to database as integer number. I have create four functions use as following:

  1. ptime to integer (by day):

     int ti=time_to_int(ptime(date(2012,1,1),0));
 => ti = 15430;//number of day from 1970 to 2012

  2. integer to ptime:

     ptime pt = int_to_time(15430);

  3. ptime to string: 

     ptime pt(date(2012,1,1), 0);
 string s = time_to_string(pt, "%d.%m.%Y");

  4. string to ptime:

    ptime pt = string_to_time(%d.%m.%Y);

PROBLEM: Above, I have used epoch from 1970 and All work very well. So how can I convert 01/01/1945 to integer? I think that need use epoch from such as 1900. However, when convert 01/01/2012 to int from 1900 it returns negative number, because "I think" tick count in miliseconds overflow (32bit). Use some ways manual to calculate date to int is ok, but convert int back to date seems to bad. I want to use boost to do this.

Any suggestion? Thanks!

Upvotes: 1

Views: 1552

Answers (1)

monex0
monex0

Reputation: 238

Integer (int) is not big enough to hold the number of seconds since 1900. It can only hold about 68 years. You need to use a long (64 bit). Boost's time_duration will give you access to the number of seconds as a long. You could also use the number of days as mentioned by @jogojapan. However, day lengths are not always the same (23-25 hours), so sometimes keeping track of things in days is more prone to error (if you are careful it should be fine though).

Take a look at some examples from the documentation. You can subtract two ptime objects to get a time_duration:

time_duration td = ptime1 - ptime2;
long secElapsed = td.total_seconds();

Upvotes: 1

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