C integer division and floor

Question

In C, is there a difference between integer division a/b and floor(a/b) where both a and b are integers? More specifically what happens during both processes?

Possible duplicate of [What is the behavior of integer division?](https://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division) — phuclv, Jul 03 '18 at 11:16

score 23 · Accepted Answer · edited Sep 02 '12 at 23:32

23

a/b does integer division. If either a or b is negative, the result depends on the compiler (rounding can go toward zero or toward negative infinity in pre-C99; in C99+, the rounding goes toward 0). The result has type int. floor(a/b) does the same division, converts the result to double, discards the (nonexistent) fractional part, and returns the result as a double.

edited Sep 02 '12 at 23:32

obataku

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answered Sep 02 '12 at 22:26

Pete Becker

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In C integer division performs truncation towards zero. This is true since C99, before it was implementation-defined. – ouah Sep 02 '12 at 22:32
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Ah, missed the `C` tag. Still, it's clear that my answer is about C++. `` – Pete Becker Sep 02 '12 at 22:34
@Mysticial the point is that `floor` doesn't round anything in this instance because `a / b` performs integer division and *then* passes it to `floor`. – obataku Sep 02 '12 at 23:34
`-4 / 3 == -1` while `floor(-4 / 3) == -2` – Geoff Armstrong Nov 01 '16 at 18:22

score 9 · Answer 2 · edited Apr 18 '17 at 07:36

floor returns a double while a / b where both a and b are integers yields an integer value.

With the correct cast the value is the same.

If typeof operator existed in C (it does not) we would have:

(typeof (a /b)) floor(a / b) == a / b

EDIT: Now if the question is: is there any difference between:

(double) (a / b)

and

floor(a / (double) b)

the answer is yes. The results differ with respect to negative values.

score 5 · Answer 3 · answered Sep 02 '12 at 23:28

5

It's possible to lose information converting from integer to floating point. Not likely with int and double, but with slight alteration:

#include <stdio.h>
#include <math.h>

int main(void)
{
    unsigned long long a = 9000000000000000003;
    unsigned long long b = 3;
    printf("a/b = %llu\n", a/b);
    printf("floor(a/b) = %f\n", floor(a/b));
    return 0;
}

Result:

a/b = 3000000000000000001
floor(a/b) = 3000000000000000000.000000

answered Sep 02 '12 at 23:28

Alan Curry

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A double can store all 32bit integer values exactly. You can always use double instead of int. It's not just unlikely to lose precision, its impossible. Your example is correct, but misleading people who don't understand the problem yet. – maxy Jun 27 '14 at 10:12

score 3 · Answer 4 · answered Jun 27 '14 at 11:01

In general, assuming that the integers are representable in both the integer and the floating-point types, there isn't a difference, but the proof is not obvious. The problem is that in floating-point, a rounding occurs in the division a/b, so that the floor function doesn't apply on the exact rational value, but on an approximate value. I had written a paper on the subject: https://www.vinc17.net/research/publi.html#Lef2005b

In short, the result I've obtained is that if a - b is exactly representable in the floating-point system, then floor(a/b), where a and b are floating-point numbers (with integer values), gives the same result as the integer division a/b.

C integer division and floor

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