[symmetricsaurus]

There’s no reason to expect the result of integer overflow to be negative, or any particular value at all. It is undefined.

So after checking that x > o there is no way for i (a product and ratio of positive numbers) to become negative.

The only reason this is surprising is that there is an expectation of wrapping on overflow. But this is simply not the behavior of the C virtual machine.

(Maybe today it makes sense to have even C wrap in a two’s-complement way? C is used in many places though, maybe there are still platforms in use that aren’t two’s-complement?)

[josephcsible]

Be careful: it's not just that the result of the multiplication won't be what you expect. The entire execution of the program becomes undefined. For example, if you had an unconditional `puts("hello");` between the multiplication and comparison lines, if the overflow happens, it would be allowed to print "goodbye" instead.

[symmetricsaurus]

Yes, and in particular the compiler can assume that you stay within the defined behavior. That's what's enabling the optimization in removing the i < 0 check.