The first form has no undefined behavior. Since p may be NULL, in general the compiler is not allowed to make that transformation unless it can prove bar never

The first form has no undefined behavior. Since p may be NULL, in general the compiler is not allowed to make that transformation unless it can prove bar never gets called with p=NULL and n=0 (which it probably can't). If the compiler has some special knowledge of the target architecture which makes the second form behave as if the first form was executed (such as if NULL is a valid address on the target architecture), it may make the transformation, but that still wouldn't cause undefined behavior, because it must behave as if it had the first form.