A problem in my c code

Question

#include <stdio.h>

int main() {
int houses;
int percentage;
scanf("%d", &houses);

percentage = (2/houses)*100;
printf(" %d", percentage);

return 0;

•why does it prints for me 0 ?!

Answer 1

In C, a/b = 0 when a<b and a,b are integers.
Try using floats instead

Answer 2

printf("%f",percentage);

If you want to get an integer output, try using ceil() or floor() as you expected.

Answer 3

What if i wanted only whole numbers in my output as percentage ?

Answer 4

Simba

#include <stdio.h>

int main() {
float houses;
float percentage;
scanf("%f", &houses);

percentage = (2/houses)*100;

printf(" %d", percentage);

return 0;
}
Now when houses = 3 it gives me percentage=1. Why ?

Answer 5

You may try typecast the division:

percentage = (float)2/houses*100

don't forget to use "%f" instead of "%d" in printf() in the final answer

Answer 6

I found two interesting options and both uses typecasting:

Integer percentage:

#include <stdio.h>
int main(void) {
int houses;
int percentage;
scanf("%d", &houses);

percentage = (float)2/houses*100;
printf("%d", percentage);
}

Float percentage:

#include <stdio.h>
int main(void) {
int houses;
float percentage;
scanf("%d", &houses);

percentage = (float)2/houses*100;
printf("%f", percentage);
}

In the second option you need only to change type of percentage (float percentage) and of course use "%f" in final printf().

Tip: for typecast you must avoid round brackets ( ) see the code.

Answer 7

If you prefer not use typecasting, define percentage as float and use 2.0 instead of 2 in the division. Hope I helped ;)

Answer 8

Thankyou.🤝