Cast double to int correctly(5 lines of code)

[tmac619619]

#define CENTS_PER_DOLLAR 100 

int avg_min_cost;
 avg_min_cost=(wkday_min+night_min+wkend_min))/(CENTS_PER_DOLLAR);

  
printf("Average minute cost: $ %.2lf\n",(double)avg_min_cost);

I'm not sure in the above code whether or not i casted the int to doubles correctly.

Here are the specifications:
"Note that since you are dividing by 100, and not 100.0
You will have to cast your int variable to doubles in your printf statements before dividing by CENTS_PER_DOLLAR"

ps: when dividing by 100.0 the program works fine. Dividing by the 100, the numbers are off from the goal.
However, i am required to divide by 100... what lol?

[Adak]

Nope.

Your instructions state to cast your int variable to doubles BEFORE dividing by CENTS_PER_DOLLAR in the printf statement.

You're not dividing in the right statement, and casting the var AFTER you've done integer division will not give you anything other than a whole number.

[tmac619619]

Nope.

Your instructions state to cast your int variable to doubles BEFORE dividing by CENTS_PER_DOLLAR in the printf statement.

You're not dividing in the right statement, and casting the var AFTER you've done integer division will not give you anything other than a whole number.

So if i do not cast my int variable into a double, and divide by 100, my result would be an integer, correct?

But if i go ahead and cast the variable, then when i divide by CENTS_PER_DOLLAR (100), it is equivalent to dividing by 100.0?

Thanks for clarifying this.

[iMalc]

It's all about what happens when you perform a binary operation on two different types, e.g addition.
One of the operands changes type first to make them the same, and the one that changes is pretty much the one that represents the least values. Look up the type promotion rules.