signed integer division with rounding in C

Question

I'd like to calculate x/y where x and y are both signed integers, and get a result rounded to the nearest integer. Specifically, I'd like a function rquotient(x, y) using integer-only arithmetic such that:

ASSERT(rquotient(59, 4) == 15);
ASSERT(rquotient(59, -4) == -15);
ASSERT(rquotient(-59, 4) == -15);
ASSERT(rquotient(-59, -4) == 15);

ASSERT(rquotient(57, 4) == 14);
ASSERT(rquotient(57, -4) == -14);
ASSERT(rquotient(-57, 4) == -14);
ASSERT(rquotient(-57, -4) == 14);

I've looked to S.O. for a solution and found the following (each with their own shortcoming):

• Rounding integer division (instead of truncating) (round up only)
• Integer division with rounding (positive x and y only)
• Round with integer division (positive x and y only)
• integer division, rounding (positive y only, but a good suggestion in the comments)
• Integer division rounding with negatives in C++ (question about the standard, not a solution)

Answer 1

If you know x and y both to be positive:

int rquotient_uu(unsigned int x, unsigned int y) {
  return (x + y/2) / y;
}

If you know y to be positive:

int rquotient_su(int x, unsigned int y) {
  if (x > 0) {
    return (x + y/2) / y;
  } else {
    return (x - y/2) / y;
  }
}

If both are signed:

int rquotient_ss(int x, int y) {
  if ((x ^ y) >= 0) {            // beware of operator precedence
    return (x + y/2) / y;        // signs match, positive quotient
  } else {
    return (x - y/2) / y;        // signs differ, negative quotient
  }
}

And if you really want to baffle your future self or are addicted to code golf, please resist the urge to write it this way: ;)

int rquotient_ss(int x, int y) {
  return (x + (((x^y)>=0)?y:-y)/2)/y;
}

Answer 2

A simple solution would be to use round and double:

#include <math.h>

int rquotient(int const x, int const y) {
    return (int)round((double)x / y);
}