What is the right way to find the average of two values?

Question

I recently learned that integer overflow is an undefined behavior in C (side question - is it also UB in C++?)

Often in C programming you need to find the average of two

Answer 1

The simplest (and usually fastest) way to average two int over the entire range [INT_MIN...INT_MAX] is to resort to a wider integer type. (Suggested by @user3100381.) Let us call that int2x.

int average_int(int a, int b) {
  return ((int2x) a + b)/2;
}

Of course this obliges a wider type to exist - so let's look at a solution that does not require a wider type.

Challenges:

Q: When one int is odd and the other is even, which way should rounding occur?
A: Follow average_int() above and round toward 0. (truncate).

Q: Can code use %?
A: With pre-C99 code, the result of a % 2 allows for different results when a < 0. So let us not use %.

Q: Does int need to have about symmetric range of positive and negative numbers?
A: Since C99 the number of negative numbers is the the same (or 1 more) than the number of positive numbers. Let us try not to require this.

SOLUTION:

Perform tests to determine is if overflow may occur. If not, simple use (a + b) / 2. Otherwise, add half the difference (signed same as answer) to the smaller value.

The following gives the same answer as average_int() without resorting to a wider integer type. It protects against int overflow and does not require INT_MIN + INT_MAX to be 0 or -1. It does not depends on encoding to be 2's complement, 1's complement or sign-magnitude.

int avgC2(int a, int b) {
  if (a >= 0) {
    if (b > (INT_MAX - a)) {
      // (a+b) > INT_MAX
      if (a >= b) {
        return (a - b) / 2 + b;
      } else {
        return (b - a) / 2 + a;
      }
    }
  } else {
    if (b < (INT_MIN - a)) {
      // (a+b) < INT_MIN
      if (a <= b) {
        return (a - b) / 2 + b;
      } else {
        return (b - a) / 2 + a;
      }
    }
  }
  return (a + b) / 2;
}

At most, 3 if()s occur with any pair of int.