Fast ceiling of an integer division in C / C++


Sparky's answer is one standard way to solve this problem, but as I also wrote in my comment, you run the risk of overflows. This can be solved by using a wider type, but what if you want to divide long longs?

Nathan Ernst's answer provides one solution, but it involves a function call, a variable declaration and a conditional, which makes it no shorter than the OPs code and probably even slower, because it is harder to optimize.

My solution is this:

q = (x % y) ? x / y + 1 : x / y;

It will be slightly faster than the OPs code, because the modulo and the division is performed using the same instruction on the processor, because the compiler can see that they are equivalent. At least gcc 4.4.1 performs this optimization with -O2 flag on x86.

In theory the compiler might inline the function call in Nathan Ernst's code and emit the same thing, but gcc didn't do that when I tested it. This might be because it would tie the compiled code to a single version of the standard library.

As a final note, none of this matters on a modern machine, except if you are in an extremely tight loop and all your data is in registers or the L1-cache. Otherwise all of these solutions will be equally fast, except for possibly Nathan Ernst's, which might be significantly slower if the function has to be fetched from main memory.


Given integer values x and y, C and C++ both return as the quotient q = x/y the floor of the floating point equivalent. I'm interested in a method of returning the ceiling instead. For example, ceil(10/5)=2 and ceil(11/5)=3.

The obvious approach involves something like:

q = x / y;
if (q * y < x) ++q;

This requires an extra comparison and multiplication; and other methods I've seen (used in fact) involve casting as a float or double. Is there a more direct method that avoids the additional multiplication (or a second division) and branch, and that also avoids casting as a floating point number?

You could use the div function in cstdlib to get the quotient & remainder in a single call and then handle the ceiling separately, like in the below

#include <cstdlib>
#include <iostream>

int div_ceil(int numerator, int denominator)
        std::div_t res = std::div(numerator, denominator);
        return res.rem ? (res.quot + 1) : res.quot;

int main(int, const char**)
        std::cout << "10 / 5 = " << div_ceil(10, 5) << std::endl;
        std::cout << "11 / 5 = " << div_ceil(11, 5) << std::endl;

        return 0;

There's a solution for both positive and negative x but only for positive y with just 1 division and without branches:

int ceil(int x, int y) {
    return x / y + (x % y > 0);

Note, if x is positive then division is towards zero, and we should add 1 if reminder is not zero.

If x is negative then division is towards zero, that's what we need, and we will not add anything because x % y is not positive