C - How to print time in format: 2009‐08‐10 18:17:54.811

Question

What's the best method to print out time in C in the format 2009‐08‐10
18:17:54.811?

Atm, I have Thu Sep 9 11:10:08 2010 but can't figure out the above.

Answer 1

Use strftime().

#include <stdio.h>
#include <time.h>

int main()
{
    time_t timer;
    char buffer[25];
    struct tm* tm_info;

    time(&timer);
    tm_info = localtime(&timer);

    strftime(buffer, 25, "%Y:%m:%d%H:%M:%S", tm_info);
    puts(buffer);

    return 0;
}

For milliseconds part, have a look at this question. How to measure time in milliseconds using ANSI C?

Answer 2

You could use strftime, but struct tm doesn't have resolution for parts of seconds. I'm not sure if that's absolutely required for your purposes.

struct tm tm;
/* Set tm to the correct time */
char s[20]; /* strlen("2009-08-10 18:17:54") + 1 */
strftime(s, 20, "%F %H:%M:%s", &tm);

Answer 3

time.h defines a strftime function which can give you a textual representation of a time_t using something like:

#include <stdio.h>
#include <time.h>
int main (void) {
    char buff[100];
    time_t now = time (0);
    strftime (buff, 100, "%Y-%m-%d %H:%M:%S.000", localtime (&now));
    printf ("%s\n", buff);
    return 0;
}

but that won't give you sub-second resolution since that's not available from a time_t. It outputs:

2010-09-09 10:08:34.000

If you're really constrained by the specs and do not want the space between the day and hour, just remove it from the format string.