How "const int *const & variable" is interpreted in c++

Question

When one aliases two variables as

int a;
const int &b = a;

the two variables are effectively the same thing, so that any change applied to the variable a is also applied to the variable b. However, when the same trick is done with pointers, it does not seem to work in the same way, as is demonstrated by the following program:

#include <iostream>
int main(void) {

    int *a = (int*) 0x1;
    const int *const &b = a;// Now b should be an alias to a.
    a = (int*) 0x2;// This should change b to 0x2.
    std::cout << b << "\n";// Outputs 0x1 instead of the expected value of 0x2.

    return 0;
}

Now the variable a does not seem to be an alias of the variable b after all, but why?

Answer 1

const int *const & is a reference to const pointer to const int. (Try to read it from right to left.) Note that the pointer's type is const int *, but not int * (i.e. the type of a). References can't be bound with different type directly. For const int *const &b = a; a temporary^* (with type const int *, copied from a) will be constructed and then bound to the reference; the temporary has nothing to do with a, so any modification on b won't effect a.

Note the difference. In the 1st sample, const is quafied on int; in the 2nd sample, const is qualified on not only the pointer itself, but also the pointee, which makes two pointers different types (int * vs. const int *). If you want to qualify const on it (which seems unnecessary for your experiment) you should only qualify it on the pointer itself, i.e. int * const &.

---

^{*The lifetime of the temporary is extended to the lifetime of reference b.}

Answer 2

const int * const & b means reference to const pointer to const int. What you want is int * const & b

Use this handy tool to decipher complex declarations. https://cdecl.org/