> The type of p is char[] which decays into a pointer and the type of (p+1) is char . Because it is char and not char * restrict, any assignment through such a

> The type of p is char[] which decays into a pointer and the type of (p+1) is char . Because it is char and not char * restrict, any assignment through such a pointer can be assumed to modify any and all data in a program. Out of bounds access is classic UB. The compiler is allowed to assume that you never access an array out of bounds. Since you never access p out of bounds, the effects of any write to p will not affect any "object" that is not p. >Therefore, the assumption that p[0] does not change and can be swapped out is not somehow surprisingly wrong, but simply entirely unfounded. I hope you notice your small mistake. It's print(q[0]), not print(p[0]), there is no "p" in the print. It's q[0] and q[0] is not p. The write to p did not affect q, because that would be UB. >It is up to the compiler to prove for itself that assignment through pointers does not change data it's interested in from before to after the assignment. (except, of course, in the case of restrict pointers, where the programmer makes this guarantee). This is not very difficult, at least conceptually. The code accesses q[0], not iq. You might say that if you did print(*iq), that the compiler should definitively know that iq and ip are the same and therefore a write to ip would also be a write to iq, but the code uses q[0], which is a memory address that is not proven to be identical to ip. Then again, what you're hoping for here is that the compiler fixes the UB for you, which is not the point of UB. You're supposed to rely on your godly C programmer instincts to never write code with UB in the first place.