[nomy99]

can you explain to me how that fixes it? Thanks

[dan-robertson]

There is an implicit invariant that 0 <= min, and min <= max, so max - min is between 0 and int_max inclusive, so you avoid overflowing with that calculation. To be sure that you avoid overflowing at the sum step, you need to prove that min + (max - min) / 2 <= max.

[nomy99]

what if max is already overflown due to size of array (without even doing the lo + high calculation?) Then min + (max- min)/2 would be min + (smaller negative value) which would violate the min + (max-min)/2 <= max

[pk86]

Assuming the value of max passed in is "sane" (non-negative and within array bounds), it is strictly decreasing and should never overflow.

[siawyoung]

This makes sure that none of the quantities being calculated will overflow while still returning the same answer.

Where x is min and y is max, the midpoint is:

(x+y)/2 = x+(y-x)/2 x+y = 2x+y-x x+y=x+y