C print issue, Decimals equating to 0 - c

Question

Im essentially brand new to C, so bear with me. So basically i have to create a program to calculate the amount of dough per sqft. After rechecking my math and debugging for a very long time; ive found that sradius/INCHES_PER_FEET is coming out to 0.
INCHES_PER_FEET = 12. Now the issue is if i input any value equal to or greater than 12 for the radius, the function will work properly; but if i input 8, and 8/12 is less than 1 a decimal, the program will automatically equate it to 0 and im not sure how i can fix that. Insight would be really appreciated.
int main(){
//define radius variables
int sradius;
printf("What is the radius of your small pizza, in inches?\n");
scanf("%d", &sradius);
int mradius;
printf("What is the radius of your medium pizza, in inches?\n");
scanf("%d", &mradius);
int lradius;
printf("What is the radius of your large pizza, in inches?\n");
scanf("%d", &lradius);
//define pizzas sold variables
int spizzas;
printf("How many small pizzas do you expect to sell this week?\n");
scanf("%d", &spizzas);
int mpizzas;
printf("How many medium pizzas do you expect to sell this week?\n");
scanf("%d", &mpizzas);
int lpizzas;
printf("How many large pizzas do you expect to sell this week?\n");
scanf("%d", &lpizzas);
//dough calculation per size
double sdough,mdough,ldough;
sdough = (PI*((sradius/INCHES_PER_FEET)*(sradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
mdough = (PI*((mradius/INCHES_PER_FEET)*(mradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
ldough = (PI*((lradius/INCHES_PER_FEET)*(lradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
//final amount of dough
double fdough;
fdough = ((sdough*spizzas)+(mdough*mpizzas)+(ldough*lpizzas));
//print statement
printf("you need to order %.3f " ,fdough);
return 0;
}

Answer 1

Cast all the radius into a double before calculating and you will get a double as the result like this:
double sdough,mdough,ldough;
sdough = (PI*(((double)sradius/INCHES_PER_FEET)*((double)sradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
mdough = (PI*(((double)mradius/INCHES_PER_FEET)*((double)mradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
ldough = (PI*(((double)lradius/INCHES_PER_FEET)*((double)lradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);

Answer 2

C typically allows you to do operations on a single "data-type". However, if you try to do such operations, C has some preset defaults. For example, in the following code, C will assign the value of a as 0.0000 and not 0.6666, even though you have declared a as float:
int main(void)
{
float a;
int b=2, c=3;
a=b/c;
printf("%f",a);
return 0;
}
This happens because b and c are declared integers. You use type-casting in such cases, i.e., force-changing the declared data-type. The above code will give you desired output if you do the following:
int main(void)
{
float a;
int b=2, c= 3;
a=(float)b/(float)c;
printf("%f",a);
return 0;
}
The answer answered by #cool-guy depicts the same, in your specific case.

Answer 3

The excessive use (IMNSHO) of parentheses is forcing the division to be integer. If you had written
x = PI * sradius / INCHES_PER_FEET * sradius / INCHES_PER_FEET * DOUGH_PER_SQFT;
instead of
x = (PI*((sradius/INCHES_PER_FEET)*(sradius/INCHES_PER_FEET))*DOUGH_PER_SQFT);
the divisions would have been done in floating point and you would get a more accurate value. Assigning that float value to an int causes truncation instead of rounding, so the accuracy could be improved by adding a half:
x = PI * sradius / INCHES_PER_FEET * sradius / INCHES_PER_FEET * DOUGH_PER_SQFT + 0.5;
Most compilers will complain about the implicit conversion to int if warning diagnostics are sufficiently enabled.
The other answers are correct in that forcing type conversion with a typecast will also get an accurate result. Or you could simply define the denominator as a float constant:
const float INCHES_PER_FEET = 12.0;
This requires each division to be done in floating point even though the numerator is an int.