Question

In a binary search implementation, obviously:

mid = (low + high)/2

can cause overflow. I have read a lot of documentation (like this) that the following prevents the problem:

mid = (low + high) >>> 1 

However, I did not see a reason why this would work. Can anyone throw some light on this?

Answer 1

There is no such thing as a “logical right shift” in C (there’s no >>> operator), so you’re probably talking about Java.

This works because low and high are presumed to be in the range 0 to 2^31-1 (assuming we’re talking about int here). The maximum possible value of low+high is no greater than than 2^32-2, and so is representable by an unsigned int (if such a thing existed in Java). Such a thing doesn’t exist in Java, so we’ve now overflowed. However, the logical shift operator >>> treats its operand as if it were unsigned, so this gives the expected result.