What is time_t ultimately a typedef to?

Question

I searched my Linux box and saw this typedef:

typedef __time_t time_t;

But I could not find the __time_t definition.

Answer 1

The answer is definitely implementation-specific. To find out definitively for your platform/compiler, just add this output somewhere in your code:

printf ("sizeof time_t is: %d\n", sizeof(time_t));

If the answer is 4 (32 bits) and your data is meant to go beyond 2038, then you have 25 years to migrate your code.

Your data will be fine if you store your data as a string, even if it's something simple like:

FILE *stream = [stream file pointer that you've opened correctly];
fprintf (stream, "%d\n", (int)time_t);

Then just read it back the same way (fread, fscanf, etc. into an int), and you have your epoch offset time. A similar workaround exists in .Net. I pass 64-bit epoch numbers between Win and Linux systems with no problem (over a communications channel). That brings up byte-ordering issues, but that's another subject.

To answer paxdiablo's query, I'd say that it printed "19100" because the program was written this way (and I admit I did this myself in the '80's):

time_t now;
struct tm local_date_time;
now = time(NULL);
// convert, then copy internal object to our object
memcpy (&local_date_time, localtime(&now), sizeof(local_date_time));
printf ("Year is: 19%02d\n", local_date_time.tm_year);

The printf statement prints the fixed string "Year is: 19" followed by a zero-padded string with the "years since 1900" (definition of tm->tm_year). In 2000, that value is 100, obviously. "%02d" pads with two zeros but does not truncate if longer than two digits.

The correct way is (change to last line only):

printf ("Year is: %d\n", local_date_time.tm_year + 1900);

New question: What's the rationale for that thinking?