Thread: float type- beginner Q

hi,

what is wrong with this code :

Code:

#include <stdio.h>
#include <stdlib.h>

int main(){
int a =1;
int b = 2;
int c = 0;
float i = 0;

c= a-b;
i = a/b;
printf("%d,%f\n",c,i);
}

I am expecting to print 2 numbers : -1 and 0.5, but i get -1 and 0.000000. What am i doing wrong

thank you

baxy

Code:

i = (float)a/(float)b;

What happens is that the division kills your precision

When you do a/b where both 'a' and 'b' are ints, the compiler uses integer arithmetic and returns an int result. If you want to ensure the division is done using floating point, force one of the numbers to be float: i = (float)a / b