Why can't we pass int array[] to hoo(int* &p)?

Question

In my understanding array in int array[]={1,2,3,4,5} is just a pointer to the first element of array. It means that array can be assigned to a pointer ptr of type int*.

Parameter int* &p in hoo will pass the argument by reference. It means we can change the passed argument to point to another value from within the hoo.

void hoo(int* &p, int n)
{
    for (int i = 0; i < n; i++)
        cout << p[i] << endl;
}

int main()
{
    int array[] = { 1,2,3,4,5 };

    // I can do this
    int* ptr = array;
    hoo(ptr, 5);

    // but not this.
    //hoo(array, 5);
}

Question

Why can't we pass int array to hoo without ptr ?

Answer 1

In my understanding array in int array[]={1,2,3,4,5} is just a pointer to the first element of array.

This is not correct. Arrays are arrays and pointers are pointers. They are distinct types with distinct properties. They are often confused because an array has the property that it will eagerly decay to a pointer to its first element.

hoo(array, 5); tries to convert array to an int* but the result of that conversion is an rvalue and can't be bound to a non-const reference. If, for example, you changed hoo to take a const reference it will compile fine :

void hoo(int* const &p, int n) { }

int main()
{
    int array[] = { 1,2,3,4,5 };
    hoo(array, 5);
}

In that case, you cannot change what p points to, making the use of a reference pointless.