It has to. Just imagine what happens when in the iterative algorithm count = count + i is just before overflowing. In that case the O(N) algorithm has no over

It has to. Just imagine what happens when in the iterative algorithm count = count + i is just before overflowing. In that case the O(N) algorithm has no overflow, but if you look at the closed form solution. Surely (count)*(count-1) would overflow. A compiler must never be allowed to swap a non-overflow with a overflow. Which is one big limit of these optimizations.