That multiplication is entirely implicit, so there is no reason the compiler needs to handle it the same as it handles an explicit multiplication. Given that `p

That multiplication is entirely implicit, so there is no reason the compiler needs to handle it the same as it handles an explicit multiplication. Given that `p + i` is obviously not an integer addition and it already has much more UB then `i + j`, there is no reason why `i + j` having defined overflow rules needs to mean `p + i` also has them (just like `i + j` is safe for any small enough i and j, while p + i is only meaningful if points within the same object as p (to be fair, its not UB to compute p + i for any i, it's UB to use the value).