指针指针等指针算法

Question

Is to well defined to use pointer arithmetic on pointers to pointers? eg

int a=some_value;
int* p=&a;
int**p2=&p;

Now would it be well defined behavior to perform arithmetic on p2?(eg p2+1, p2+2,etc)

Answer 1

The other answer is completely wrong:

int *p = &a;

declares a single pointer variable (not a array of pointers). This is equivalent, WRT indexing, with an array with a single element:

int *(array[1]) = { &a };

so you can do (&array[0]) + 1 or array + 1 , or p + 1 , because forming a one-past-the-end pointer is allowed: a one-past-the-end pointer points to an imaginary element that's just after the end of the array. The one-past-the-end pointer is not dereferenceable, because it points to no real object.

But you cannot compute any other pointer value .

In particular, p+2 is not valid.

Answer 2

Of course!

p + n

where p is a pointer and n is an integer is always well-defined. It produces the address which is "n times the size of the element type p points to" bytes from p itself. In this case p2 is a pointer to a pointer. So p2 + 4 is the address "4 * the-size-of-pointers" bytes past p2 .

Since you are pointing to local variables in your specific example, it would be odd. But it will not be illegal.