Question

When x is a L-value (let’s say a variable), then the following identity holds:

x == *(&x)

This is quite easy to explain, because &x is a pointer to x and the dereference operator * applied to &x will then of course return x.

Now I am wondering if the converse makes sense. To be precise I am wondering if

p == &(*p)

when p is a non-dangling pointer.
This seems to make sense, because *p is itself a L-value (a value which has an adress), because we already have the pointer (=adress) p to it. So you only need to know, that such pointers are unique, because then &(*p) has no other chance as to be p.

So when both identities are true you can say, that, mathematically, * and & are inverse functions of one another.

Am I correct? Are there any possible exceptions to this alleged rules?

Answer 1

When p is an object pointer, &*p is equivalent to p. The evaluation of *p doesn’t occur and this is guaranteed by the C Standard.

char *q, *p = NULL;
q = &*p;  // equivalent to q = p;

Here is the relevant paragraph of the Standard:

(C99, 6.5.3.2p3) “If the operand is the result of a unary * operator, neither that operator nor the & operator is evaluated and the result is as if both were omitted, except that the constraints on the operators still apply and the result is not an lvalue.”

EDIT: After @ldav1s comment, I changed the word pointer to object pointer. Indeed if p is of type void *, then &*p is invalid. For information, this has been discussed by the C Committee in Defect Report #102: http://www.open-std.org/jtc1/sc22/wg14/www/docs/dr_012.html