Rounding integer division (instead of truncating)

Question

I was curious to know how I can round a number to the nearest whole number. For instance, if I had:

int a = 59 / 4; 

which would be 14.75 if calculated in floating point; how can I store the result as 15 in "a"?

Answer 1

The standard idiom for integer rounding up is:

int a = (59 + (4 - 1)) / 4; 

You add the divisor minus one to the dividend.

Answer 2

A code that works for any sign in dividend and divisor:

int divRoundClosest(const int n, const int d) {   return ((n < 0) ^ (d < 0)) ? ((n - d/2)/d) : ((n + d/2)/d); } 

In response to a comment "Why is this actually working?", we can break this apart. First, observe that n/d would be the quotient, but it is truncated towards zero, not rounded. You get a rounded result if you add half of the denominator to the numerator before dividing, but only if numerator and denominator have the same sign. If the signs differ, you must subtract half of the denominator before dividing. Putting all that together:

(n < 0) is false (zero) if n is non-negative (d < 0) is false (zero) if d is non-negative ((n < 0) ^ (d < 0)) is true if n and d have opposite signs (n + d/2)/d is the rounded quotient when n and d have the same sign (n - d/2)/d is the rounded quotient when n and d have opposite signs 

If you prefer a macro:

#define DIV_ROUND_CLOSEST(n, d) ((((n) < 0) ^ ((d) < 0)) ? (((n) - (d)/2)/(d)) : (((n) + (d)/2)/(d))) 

The linux kernel macro DIV_ROUND_CLOSEST doesn't work for negative divisors!

EDIT: This will work without overflow:

int divRoundClosest( int A, int B ) { if(A<0)     if(B<0)         return (A + (-B+1)/2) / B + 1;     else         return (A + ( B+1)/2) / B - 1; else     if(B<0)         return (A - (-B+1)/2) / B - 1;     else         return (A - ( B+1)/2) / B + 1; }