Extracting and formatting today's date

I would like to display today's date in the format dd/mm/yyyy. What is the best function to use to do so? I know that the time() function will give me number of seconds since Jan.1,1970, but do I have to parse the rest of the info. out from that or is there an easier way/function to use??

you can use the function asctime(const struct tm* pointer)

This returns a pointer to a string which contains the information that you need ( and a little more besides). You then have to break that string down a little to do what you want.

cant remember the exact format of the string, its something like

day/date/month/time/year /0 or something similar. look it up in your compiler helpfiles.

Try this:

int tm_year;
struct tm *systime;
time_t t;
t=time(NULL);
systime=localtime(&t);

if(systime->tm_hour>12)
{systime->tm_hour=systime->tm_hour-12; }
///so output isn't military format///

printf("Today is %.2d/ %.2d/ %.2d !!\n ",
systime->tm_mon+1,
systime->tm_mday,
systime->tm_year+1900);

///"systime" is a structure defined in time.h

:)
Thanx Sebastiani, for helping me see the light!

#include <stdio.h>
#include <time.h>
int main() {
time_t t;

time(&t);
printf(ctime(&t));

system("PAUSE");
return 0;
}



Sall