[pascal_cuoq]

Oh, that was your question. In this case, the reason why &a + 1 == &b is unspecified is that:

- it's generally false—there is no reason for b to be just after a in memory, so these two addresses compare different.

- it is sometimes true: when addresses are implemented as integers, and compilers use exactly sizeof(T) bytes to represent an object of type T, and do not waste precious integers by leaving gaps between objects, and == between pointers is implemented as the assembly instruction that compares integers, sometimes that instruction produces true for &a + 1 == &b, because b was placed just after a in memory.

In short, &a + 1 == &b was made unspecified so that compilers could implement pointer == by the integer equality instruction, and could place objects in memory without having to leave gaps between them. Anything more specific (such as “&a + 1 == &b is always false”) would have forced compilers to take additional measures against providing the wrong answer.