Is incrementing a null pointer well-defined?

Question

There are lots of examples of undefined/unspecified behavior when doing pointer arithmetics - pointers have to point inside the same array (or one past the end), or inside t

Answer 1

As said by Columbo it is UB. And from a language lawyer point of view this is the definitive answer.

However all C++ compiler implementations I know will give same result :

int *p = 0;
intptr_t ip = (intptr_t) p + 1;

cout << ip - sizeof(int) << endl;

gives 0, meaning that p has value 4 on a 32 bit implementation and 8 on a 64 bits one

Said differently :

int *p = 0;
intptr_t ip = (intptr_t) p; // well defined behaviour
ip += sizeof(int); // integer addition : well defined behaviour 
int *p2 = (int *) ip;      // formally UB
p++;               // formally UB
assert ( p2 == p) ;  // works on all major implementation