Calculating mid in binary search

Question

I was reading an algorithms book which had the following algorithm for binary search:

public class BinSearch {
  static int search ( int [ ] A, int K ) {
    int l = 0 ;
    int u = A. length −1;
    int m;
    while (l <= u ) {
      m = (l+u) /2;
      if (A[m] < K) {
        l = m + 1 ;
      } else if (A[m] == K) {
        return m;
        } else {
          u = m−1;
        }
       }
       return −1;
      }
 }

The author says "The error is in the assignment m = (l+u)/2; it can lead to overflow and should be replaced by m = l + (u-l)/2."

I can't see how that would cause an overflow. When I run the algorithm in my mind for a few different inputs, I don't see the mid's value going out of the array index.

So, in which cases would the overflow occur?

Answer 1

This post covers this famous bug in a lot of detail. As others have said it's an overflow issue. The fix recommended on the link is as follows:

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

// Alternatively
int mid = (low + high) >>> 1;

It is also probably worth mentioning that in case negative indices are allowed, or perhaps it's not even an array that's being searched (for example, searching for a value in some integer range satisfying some condition), the code above may not be correct as well. In this case, something as ugly as

(low < 0 && high > 0) ? (low + high) / 2 : low + (high - low) / 2

may be necessary. One good example is searching for the median in an unsorted array without modifying it or using additional space by simply performing a binary search on the whole Integer.MIN_VALUE–Integer.MAX_VALUE range.