How can I safely average two unsigned ints in C++?

Question

Using integer math alone, I\'d like to \"safely\" average two unsigned ints in C++.

What I mean by \"safely\" is avoiding overflows (and anything else that can be th

Answer 1

What you have is fine, with the minor detail that it will claim that the average of 3 and 3 is 2. I'm guessing that you don't want that; fortunately, there's an easy fix:

unsigned int average = a/2 + b/2 + (a & b & 1);

This just bumps the average back up in the case that both divisions were truncated.

Answer 2

In C++20, you can use std::midpoint:

template <class T>
constexpr T midpoint(T a, T b) noexcept;

The paper P0811R3 that introduced std::midpoint recommended this snippet (slightly adopted to work with C++11):

#include <type_traits>

template <typename Integer>
constexpr Integer midpoint(Integer a, Integer b) noexcept {
  using U = std::make_unsigned<Integer>::type;
  return a>b ? a-(U(a)-b)/2 : a+(U(b)-a)/2;
}

For completeness, here is the unmodified C++20 implementation from the paper:

constexpr Integer midpoint(Integer a, Integer b) noexcept {
  using U = make_unsigned_t<Integer>;
  return a>b ? a-(U(a)-b)/2 : a+(U(b)-a)/2;
}

Answer 3

And the correct answer is...

(A&B)+((A^B)>>1)

Answer 4

Your method is not correct if both numbers are odd eg 5 and 7, average is 6 but your method #3 returns 5.

Try this:

average = (a>>1) + (b>>1) + (a & b & 1)

with math operators only:

average = a/2 + b/2 + (a%2) * (b%2)