What I meant to say is that if you have an overflow in those cases, the results of the expression is definitely not what you wanted - this is going to propagate

What I meant to say is that if you have an overflow in those cases, the results of the expression is definitely not what you wanted - this is going to propagate and "damage" other expressions. Applying the optimization in that case might produce a different result. An example is this new transformation from the blog post: (a * C1) / C2 -> a * (C1/C2), where a * C1 might overflow. For 8 bit numbers with a = 106 and C1 = C2 = 17 we get initial: (106 * 17) / 17 = 10 (overflow) / 17 = 0 optimized: 106 * (17 / 17) = 106 * 1 = 106 So in this case the optimized version gives the expected result - it's still different than the initial expression, so it falls under the "undefined overflow" optimizations category.