[pksadiq]

  int mid(int x, int y) {
    return (x/2 + y/2) + (1 & x & y);
  }

would be a more readable solution

edit: Actually, this fails on mid(INT_MIN, INT_MAX) and possibly other mixed sign values (returns: -1, expected: 0 (or -1 is okay?), where the precise answer is -0.5)

more edit: The C standard says: When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded (This is often called ‘‘truncation toward zero’’).

So -1/2 should be 0.

[nwellnhof]

For signed ints, try

    (x >> 1) + (y >> 1) + (x & y & 1)

This rounds toward negative infinity. Also

    (x >> 1) + (y >> 1) + ((x | y) & 1)  // Rounds toward +Inf
    (x >> 1) + (y >> 1) + (x & 1)        // Rounds up if x is odd
    (x >> 1) + (y >> 1) + (y & 1)        // Rounds up if y is odd

I'd be curious if there's a bit-twiddling trick for rounding toward x or y.

[tiffanyh]

Using the midpoint function is the most readable.

https://en.cppreference.com/w/cpp/numeric/midpoint

Though it's odd this allows overflow.

[abainbridge]

std::midpoint also seems to yield less efficient code with g++12 targeting x86-64: https://godbolt.org/z/j695ce98Y

[Lockal]

std::midpoint does different thing in terms of rounding, therefore comparison is not fair.

[abainbridge]

Possibly true, but it yields less efficient code in GCC 12 for x86-64. https://godbolt.org/z/oafPrb4K8

[bonzini]

Because it's also wrong! / rounds towards zero, while summing a&b&1 requires the division to round towards negative infinity.