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

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

-------------Problems Reply------------

Use strftime().

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

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

time(&timer);
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?

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.

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);

Category:c# Views:0 Time:2010-09-09
Tags: c# time

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