Reference is not an object.

[wy]

I'm reading "C++ Primer", and it emphasizes "reference is not an object".
In P51, it says "Because references are not objects, we may not define a
reference to a reference." And in P52, it says "Because references are
not objects, they don't have addresses. Hence, we may not define a
pointer to a reference."

But the following code works with g++.

#include <iostream>

using namespace std;

int main()
{
int val = 0xfe;
int &ref= val;
int &r = ref;
int *p = &ref;
cout << "val = " << val << endl;
cout << "ref = " << ref << endl;
cout << "r = " << r << endl;
cout << "*p = " << *p << endl;
return 0;
}

Is the feature not standard, or do I misunderstand?

[Leigh Johnston]

On 11/11/2013 01:08, wy wrote:
> I'm reading "C++ Primer", and it emphasizes "reference is not an object".
> In P51, it says "Because references are not objects, we may not define a
> reference to a reference." And in P52, it says "Because references are
> not objects, they don't have addresses. Hence, we may not define a
> pointer to a reference."
>
> But the following code works with g++.
>
> #include <iostream>
>
> using namespace std;

Don't do "using namespace std;" as it is amateurish.

>
> int main()
> {
> int val = 0xfe;
> int &ref= val;
> int &r = ref;
> int *p = &ref;
> cout << "val = " << val << endl;
> cout << "ref = " << ref << endl;
> cout << "r = " << r << endl;
> cout << "*p = " << *p << endl;
> return 0;
> }
>
> Is the feature not standard, or do I misunderstand?

You are misunderstanding the code above. Nowhere in that code is a
pointer to a reference being made; 'p' is the address of 'val' not the
address of the reference to 'val'.

/Leigh

[Ian Collins]

You misunderstand!

int &r = ref;

doesn't declare a reference to a reference, it declares another
reference that is assigned the same value as ref.

int *p = &ref;

declares an int pointer to the address of whatever ref is bound to. Yo
can see this by adding the line

cout << "p = " << p << " &val = " << &val << endl;

int& *p;

would attempt to declare a pointer to a reference.

--
Ian Collins

[wy]

On 11/11/2013 09:18 AM, Ian Collins wrote:
> int &r = ref;
>
> doesn't declare a reference to a reference, it declares another
> reference that is assigned the same value as ref.

I say this on P69 of the book. "When we use a reference, we are really
using the object to which the reference refers." Thank you!

> int& *p;
>
> would attempt to declare a pointer to a reference.
>

So "int &&" attempts to declare a reference to a reference, "int &*"
attempts to declare a pointer to a reference, right?

[Öö Tiib]

No. "int &&" is reference to int rvalue. "int &*" is syntax error
that looks like someone wanted to declare pointer to reference.

[sg]

Am 12.11.2013 02:23, schrieb wy:
> So "int &&" attempts to declare a reference to a reference,

No. Careful. && is a single token. What you should have written is:

int & &

with a space in between the ampersands. But since this doesn't work
anyways, you can just forget about it. :)

Starting with C++11 the double-ampersand (as single token) can also be
used to declare a reference. It's another kind of reference:

int & lvalue reference to int
int && rvalue reference to int (NOT a ref to a ref!)

> "int &*" attempts to declare a pointer to a reference, right?

Yes. But this also does not work, because a reference it not an object.
You can only create pointers to objects or pointers to functions.