Why dividing two integers doesn't get a float? [duplicate]
Solution 1:
This is because of implicit conversion. The variables b, c, d
are of float
type. But the /
operator sees two integers it has to divide and hence returns an integer in the result which gets implicitly converted to a float
by the addition of a decimal point. If you want float divisions, try making the two operands to the /
floats. Like follows.
#include <stdio.h>
int main() {
int a;
float b, c, d;
a = 750;
b = a / 350.0f;
c = 750;
d = c / 350;
printf("%.2f %.2f", b, d);
// output: 2.14 2.14
return 0;
}
Solution 2:
Use casting of types:
int main() {
int a;
float b, c, d;
a = 750;
b = a / (float)350;
c = 750;
d = c / (float)350;
printf("%.2f %.2f", b, d);
// output: 2.14 2.14
}
This is another way to solve that:
int main() {
int a;
float b, c, d;
a = 750;
b = a / 350.0; //if you use 'a / 350' here,
//then it is a division of integers,
//so the result will be an integer
c = 750;
d = c / 350;
printf("%.2f %.2f", b, d);
// output: 2.14 2.14
}
However, in both cases you are telling the compiler that 350 is a float, and not an integer. Consequently, the result of the division will be a float, and not an integer.
Solution 3:
"a" is an integer, when divided with integer it gives you an integer. Then it is assigned to "b" as an integer and becomes a float.
You should do it like this
b = a / 350.0;
Solution 4:
Specifically, this is not rounding your result, it's truncating toward zero. So if you divide -3/2, you'll get -1 and not -2. Welcome to integral math! Back before CPUs could do floating point operations or the advent of math co-processors, we did everything with integral math. Even though there were libraries for floating point math, they were too expensive (in CPU instructions) for general purpose, so we used a 16 bit value for the whole portion of a number and another 16 value for the fraction.
EDIT: my answer makes me think of the classic old man saying "when I was your age..."