C++ -- Why do we not dereference when passing by reference?

I recently got back into trying to teach myself C++, mostly through YouTube tutorials (this guy's are pretty good). I only went to IRC for an answer at one point, but would like to share something I found & wasn't able to Google on.

Everyone should know how to use pointers when working with C++.. say we had:

using namespace std;

int main()
{
    int num = 5;
    int *pnum = #
    cout<<"integer num (@"<<<") = "<<*pnum;
    return 0;
}

Then we can see that integer num is set equal to five, and integer pointer pnum is set to the address of num. Normally, in C++, the & operator refers to the address of the object the operator is being applied to.

In the case of passing values to function by reference, I was quite confused when I found that we do not in fact need to dereference the reference passed as an argument. For example:

using namespace std;

void increment_variable(int & x)
{
        x++;
        return;
}

int main()
{
        int num = 5;

        increment_variable(num);

        cout<<"integer num (@"<<&num<<") = "<<
        return 0;
}

In the increment_variable(int & x) function, we would think that to increment the value of the x variable, we would need to dereference the address of the x variable, which has been passed as a parameter. As shown by the code, this is not in fact the case!

In reality, the & operator when used with pointers is a completely different operator than the & used when passing values to functions by reference.

Do not get the two confused!