> Yes, detecting the overflow is free but reacting to it is expensive. And if you tell the compiler not to care, it won't be doing that reaction (only the 'fre

> Yes, detecting the overflow is free but reacting to it is expensive. And if you tell the compiler not to care, it won't be doing that reaction (only the 'free' detection) and I thought you already stipulated that your hypothetical code would be doing math where that simply didn't happen, meaning that the expensive reaction would not be triggered.