Average pay in euros and cent

Hey Folks,
I did a sample exercise in which to calculate the average hourly pay rate. Asked the user to enter the weekly pay (in euros as €12.51) and the weekly hours. These *had* to be read in as floating points. The output *had* to be in euro and cent separately.

Could you guys comment on whether the technique below is elegant enough or is cumbersome? This was really a test of playing with conversions and casts.
Just look at the *internal calculations* section and say whether it could be better done. I'm fairly happy with it - I'm not interesting trapping all cases at the moment - but want to see if this is the correct approach to get the cents and euros.
Again I really appreciate your kind comments. :)

Code:

---


int main(void)
{
  double weekly_pay = 0.0;
  double weekly_hours = 0.0;
  double average_pay_per_hour = 0.0;
  int euros = 0;       
  int cents = 0;

  /* prompt user for weekly pay in euro and weekly hours worked */
  printf("\nThis program will calculate your average hourly pay rate.");
  printf("\nPlease enter your weekly pay in euro: ");
  scanf("%lf", &weekly_pay);

  printf("Please enter your weekly hours worked: ");
  scanf("%lf", &weekly_hours);

  /* internal calculations */
  average_pay_per_hour = weekly_pay / weekly_hours;     
  euros = (int)average_pay_per_hour;                      /* keep the euro, knock the cent */
  cents = (int)(average_pay_per_hour * 100) - euros*100;  /* get cents */

  /* display results */
  printf("\nYour average hourly pay rate is %d euro and %d cents.\n",euros, cents);
  return 0;
}

/*
Sample output:
This program will calculate your average hourly pay rate.
Please enter your weekly pay in euro: 105.1
Please enter your weekly hours worked: 10

Your average hourly pay rate is 10 euro and 50 cents

---

Code:

---

  euros = (int)average_pay_per_hour;                      /* keep the euro, knock the cent */
  cents = (int)(average_pay_per_hour * 100) - euros*100;  /* get cents */

---

The first line is good. But the second line is like saying

Code:

---

cents = (int)(average_pay_per_hour * 100) - ((int)average_pay_per_hour)*100;

---

which is somewhat inefficient. I don't see what's wrong with just plain

Code:

---

cents = (int)(average_pay_per_hour * 100) % 100;

---

Excellent code BTW. I couldn't have done better myself, assuming you accidently left out the closing comment symbol at the end of the file and the header files (well, just one: stdio.h). :) And you could read lines with fgets() and parse them from there and maybe check the return values of scanf() . . ..

Thanks dwks, I knew I a modulus was due to placed in there! / and % are always close together . ;) Yeah, thanks for that. It's cleaner and % is seen in many previous examples. I haven't done this in years but it's nice to get back into it. My thoughts at this stage are - "Oh my! I'm actually putting more thought into the var types and the manipulation of them rather the staightforward process of division." In Oracle, I rarely ever think about vars anymore - just NUMBER and VARCHAR2(X). But C is fun, and I can't wait to get back to the maze.c program from years back.

Code:

---

  /* internal calculations */
  average_pay_per_hour = weekly_pay / weekly_hours;
  euros = (int)average_pay_per_hour;                      /* keep the euro, knock the cent */
  cents = (int)(average_pay_per_hour * 100) % 100;        /* get cents */
  /* display results */
  printf("\nYour average hourly pay rate is %d euro and %d cents.\n",euros, cents);

---