为什么对数组的const左值引用不能在下面编译？

Question

The const lvalue reference int(const& crb)[3] = b; doesn't compile. Why?

#include<iostream>

int a = 1;
int b[3] = { 1, 2, 3 };

int main(){
    int& ra = a;             // Ok
    int const& cra = a;      // Ok

    int(&rb)[3] = b;         // Ok
    int(const& crb)[3] = b;   // Doesn't compile
}

Error message emitted by g++:

error: expected primary-expression before 'int'
int(const& crb)[3] = b;
^~~

Answer 1

Reference is always constant, you can't change reference. int const (& crb)[3] = b; here we have reference to array of const int, we can also write const int (& crb)[3] = b; It would be the same.

Pointers have a difference, pointer can be changed

const int *p; - here it is pointer on const int

int const *p; - here it is const pointer on int

const int const *p; - here it is const pointer on const int