Explanation of the safe average of two numbers

Question

Whenever I need to average two numbers for an algorithm like binary search, I always do something like this:

int mid = low + ((high - low) / 2);

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Answer 1

The code you saw is broken: it doesn't compute the average of negative numbers correctly. If you are operating on non-negative values only, like indexes, that's OK, but it's not a general replacement. The code you have originally,

int mid = low + ((high - low) / 2);

isn't safe from overflow either because the difference high - low may overflow the range for signed integers. Again, if you only work with non-negative integers it's fine.

Using the fact that A+B = 2*(A&B) + A^B we can compute the average of two integers without overflow like this:

int mid = (high&low) + (high^low)/2;

You can compute the division by 2 using a bit shift, but keep in mind the two are not the same: the division rounds towards 0 while the bit shift always rounds down.

int mid = (high&low) + ((high^low)>>1);