Question

I’d like to know if, in the following declaration, p is an lvalue, i.e. is p an expression in this declaration ? (if the answer is “yes”, it’s an lvalue because p changes the value of the object located at &p with the value &n).

unsigned char *p = &n;

Answer 1

C says (emphasis mine) for initialization of a scalar:

(C99, 6.7.8p11) “The initial value of the object is that of the expression (after conversion); the same type constraints and conversions as for simple assignment apply, taking the type of the scalar to be the unqualified version of its declared type.”

and the constraints of the assignment operators says the left operand must be a (modifiable) lvalue.

And for the terminology, a declaration is not an expression nor a statement but the initializer can be an expression.