Why dividing two integers doesn't get a float?

Question

Can anyone explain why b gets rounded off here when I divide it by an integer although it's a float?

#include <stdio.h>

void main() {
    int a;
    float b, c, d;
    a = 750;
    b = a / 350;
    c = 750;
    d = c / 350;
    printf("%.2f %.2f", b, d);
    // output: 2.00 2.14
}

http://codepad.org/j1pckw0y

Answer 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;
}

Answer 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.

Answer 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;

Answer 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..."

Answer 5

Chapter and verse

6.5.5 Multiplicative operators
...
6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.¹⁰⁵⁾ If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a; otherwise, the behavior of both a/b and a%b is undefined.

---

^{105) This is often called ‘‘truncation toward zero’’.}

Dividing an integer by an integer gives an integer result. 1/2 yields 0; assigning this result to a floating-point variable gives 0.0. To get a floating-point result, at least one of the operands must be a floating-point type. b = a / 350.0f; should give you the result you want.

Answer 6

Probably the best reason is because 0xfffffffffffffff/15 would give you a horribly wrong answer...

Answer 7

Dividing two integers will result in an integer (whole number) result.

You need to cast one number as a float, or add a decimal to one of the numbers, like a/350.0.