Algo for create hours and minutes with given seconds

Hi...this (semplified) code wait till I press a key, then it looks for the seconds elapsed.

But I want to divide them into hours/minute/seconds and not only seconds.. the algo I've wrote it's wrong (but it seems ok to me )

... can u help me ?

Code:

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

int main ()

{
  time_t start,end, timer;
  char Input [256];
  double dif;
  int ore;  //hours
  int minuti; //minutes
  int secondi;  //seconds

  timer=time(NULL);
  time (&start);
  printf("Workunit Started   --> %s",asctime(localtime(&timer)));
  gets (Input);
  time (&end);
  dif = difftime (end,start);
  timer=time(NULL);
  printf("Workunit Completed --> %s\n",asctime(localtime(&timer)));
  
  // that's wrong !!! 
  ore = dif/216000;
  minuti = (dif-(ore*216000))/3600;
  secondi = dif-((ore*216000)-(minuti*3600)); 
  
  printf ("Time elapsed : %d:%d:%d \n", ore,minuti,secondi );
  
  system("PAUSE");
 
  return 0;
}

thanx for your attention

If your algorithm doesn't work, try it using a pen and paper.

Doing an average system may help.

Especially for the minutes etc... It's not that hard of an algorithm, it just takes a while to write and understand.

Though, jotting it down may help [which I haven't myself] I like playing around with code until I understand how it works and how I did it. I can always give some recommendations of how this code works if you like.


- Stack Overflow

What is the purpose of timer in your code? It doesn't do anything, and you never actually use it. Why not just:

Code:

        start = time( NULL );
        /* do stuff */
        end = time( NULL );
        difference = difftime( end, start );
        /* calculate */
        /* display */

Quzah.

ok i'm trying to figure out why it doesnt work

ps: for Quzah... I need timer in the code I've cut (I preferred to not post a very long code)

Code:

somenumber = passednumberofseconds
somenumber % 60 = the "left over" seconds, this should be stored in a seconds counter
somenumber / 60 = number of minutes
number of minutes % 60 = "left over minutes", store this in some minutes counter
number of minutes / 60 = hours

This is like the "make correct change" exercise people commonly do. You are in effect "making correct change" from a given number of seconds (cents).

Quzah.

thanx to all

anyway I got it, it was a sign mistake, a - for a +

this works

Code:

  ore = dif/3600;
  minuti = (dif-(ore*3600))/60;
  secondi = dif-((ore*3600)+(minuti*60));