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

Use strftime().

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

int main()
{
    time_t timer;
    char buffer[26];
    struct tm* tm_info;

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

    strftime(buffer, 26, "%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?

The above answers do not fully answer the question (specifically the millisec part). My solution to this is to use gettimeofday before strftime. Note the care to avoid rounding millisec to "1000". This is based on Hamid Nazari's answer.

#include <stdio.h>
#include <sys/time.h>
#include <time.h>
#include <math.h>

int main() {
  char buffer[26];
  int millisec;
  struct tm* tm_info;
  struct timeval tv;

  gettimeofday(&tv, NULL);

  millisec = lrint(tv.tv_usec/1000.0); // Round to nearest millisec
  if (millisec>=1000) { // Allow for rounding up to nearest second
    millisec -=1000;
    tv.tv_sec++;
  }

  tm_info = localtime(&tv.tv_sec);

  strftime(buffer, 26, "%Y:%m:%d %H:%M:%S", tm_info);
  printf("%s.%03d\n", buffer, millisec);

  return 0;
}

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.