c – Implementing banker’s rounding for integer division

If the result is non-negative (a and b have same sign, i.e. both are non-negative or both are negative), then the normally rounded division result is

  x = (a + b/2) / b;

where an exact half is rounded towards +infinity.

If the result is negative, then the normally rounded division result is:

  x = (a - b/2) / b;

where an exact half is rounded towards -infinity.

This is equal to the result of “banker’s rounding” except if x is odd and the rounding was from an exact half. An exact half is obtained if “a/b=x-½” or equivalently, “b * x = a + b/2” (in the “non-negative” case) or “b * x = a – b/2” (in the “negative” case) and if that happens, then x must be decremented/incremented by 1 to get the correct result.

Notice that if b is odd, then with a and b being integers, it is not possible for “a/b” to have a fractional part of exactly ½. So the intermediate result only may need correction if b is even and in that case, the conditions above involve only integer terms.

All in all, we have:

int divround(int a, int b)
{
    int x;
    if((a<0) == (b<0))
    {   // a and b have same sign
        x = (a + b/2) / b;
        if((b%2 == 0) && (x%2 != 0) && (b*x == a + b/2))  // b even, x odd and a/b=x-1/2
        {
            x--;
        }
    }
    else
    {   // a and b have different sign
        x = (a - b/2) / b;
        if((b%2 == 0) && (x%2 != 0) && (b*x == a - b/2))  // b even, x odd and a/b=x+1/2
        {
            x++;
        }
    }
    return x;
}

Of course, if a and b are required to be non-negative, then only the first case in the if–else-statement is needed.

A few test cases:

divround(3, 2) == 2
divround(-3, 2) == -2
divround(3, -2) == -2
divround(-3, -2) == 2
divround(5, 2) == 2
divround(-5, 2) == -2
divround(5, -2) == -2
divround(-5, -2) == 2
divround(5, 1) == 5
divround(5, -1) == -5
divround(5, 3) == 2
divround(5, -3) == -2
divround(0, 2) == 0
divround(0, -2) == 0