Why are binary search index computed this way?

Question

In every code piece I read online or on book, say if someone wants to compute the midpoint between s and e, they do:

int mid = s + ((e - s) / 2);

Mathematically isn't this the same thing as

int mid = (s + e) / 2;

So why is it written in the first way often? My guess is to prevent integer overflow but not sure.

Thanks

You've gotten the right answer. In fact, some code does use `(s+e)/2`, but yes, it can lead to problems with integer overflow. — Jerry Coffin, Aug 14 '16 at 14:27
They are not the same due to how division of integers occurs. — Ed Heal, Aug 14 '16 at 14:29
@EdHeal The problem is just overflow. The rounding in the division works out the same as long as `s` and `e` are positive. — Matt Timmermans, Aug 14 '16 at 14:35
@EdHeal: Can you specify some numbers (that don't lead to overflow) that produce different results? — Jerry Coffin, Aug 14 '16 at 14:36

score 1 · Accepted Answer · answered Aug 14 '16 at 14:33

1

If e is close to the maximum value for an integer, then (s+e)/2 can overflow, but s+(e-s)/2 cannot (assuming that s is nonnegative).

For example (MAX_INT-2 + MAX_INT) == -4, so (MAX_INT-2 + MAX_INT)/2 == -2

answered Aug 14 '16 at 14:33

Matt Timmermans

1 Answers1