LinuxQuestions.org - How to print out "n" days ago's year, month and day?

I was just going to write what ta0kira wrote. The GNU date command is awesome. Anyway, in C, you can do

Code:

#include <stdio.h>

#include <time.h>



int main(void)

{

    time_t    now, then;

    struct tm *split;



    /* Get the current time */

    now = time(NULL);



    /* Convert the time to broken-down format */

    split = localtime(&now);



    /* Substract two days -- it is okay to go out of valid range */

    split->tm_mday -= 2;



    /* Renormalize the broken-down format, and return the time then. */

    then = mktime(split);

    

    printf("Two days ago was %04d-%02d-%02d, or %s\n",

          split->tm_year + 1900, split->tm_mon + 1, split->tm_mday,

          ctime(&then));



    return 0;

}

This uses the local time as a basis. If you need UTC, use gmtime instead of locatime.

The call to mktime both normalizes the split date, and returns the resulting time in the base time_t format, seconds since epoch. See the ctime, localtime, and mktime man page for details. This is in the POSIX standard, and should work as-is on all sane operating systems.

It shouldn't be too difficult to add parameter parsing to the above, to get exactly what you need.

Nominal Animal