Is "reference" a (different?) type?

[Neelesh Bodas]

Just wondering about exact terminology used by the standard to
describe a reference. More specifically, is "reference" a type?

int i = 10; // type of i is int

int &ri = i; // ri is declared as a "reference to int". But what is
type of ri? 'int' or 'reference to int'?

-Neelesh

[Thorns]

Re: Is "reference " a (different?) type?

On Jul 14, 9:57 pm, Neelesh Bodas <neelesh.bo...@ gmail.comwrote:

Just wondering about exact terminology used by the standard to
describe a reference. More specifically, is "reference" a type?
>
int i = 10; // type of i is int
>
int &ri = i; // ri is declared as a "reference to int". But what is
type of ri? 'int' or 'reference to int'?
>
-Neelesh

I believe it is 'reference to int', just like with int *p, where p is
a 'pointer to int'.
And it is a data type, because you couldn't use a int as a function's
argument, if in fact it requires, by declaration, a reference to
<data_typeor a pointer to <data_type>.

[Neelesh Bodas]

Re: Is &quot;reference &quot; a (different?) type?

On Jul 15, 12:11 am, Thorns <brokenth...@gm ail.comwrote:

>
I believe it is 'reference to int', just like with int *p, where p is
a 'pointer to int'.
And it is a data type, because you couldn't use a int as a function's
argument, if in fact it requires, by declaration, a reference to
<data_typeor a pointer to <data_type>.

That is not correct.

void f(int&); // expects a "reference to int"
void g(const int&) // expects a "reference to const int"

int main()
{
int s = 10;
f(s); // works.
g(10); // works
g(s); // works
f(10); // doesnot work for altogether different reason.
}


-N

[Kai-Uwe Bux]

Re: Is &quot;reference &quot; a (different?) type?

Neelesh Bodas wrote:

Just wondering about exact terminology used by the standard to
describe a reference. More specifically, is "reference" a type?
>
int i = 10; // type of i is int
>
int &ri = i; // ri is declared as a "reference to int". But what is
type of ri? 'int' or 'reference to int'?

Tricky. Let's see how the use of references influences overload resolution
and the typeid() operator:

#include <iostream>
#include <ostream>
#include <typeinfo>

template < typename T >
struct type_name {

static
char const * value ( void ) {
return ( "unknown" );
}

};

template <>
struct type_name<int{

static
char const * value ( void ) {
return ( "int" );
}

};

template <>
struct type_name<int&{

static
char const * value ( void ) {
return ( "int&" );
}

};

template < typename T >
const char * typeof ( T ) {
return ( type_name<T>::v alue() );
}

int& int_ref ( void ) {
static int i = 5;
return ( i );
}


#define SHOW(expr) std::cout << #expr << " = " << (expr) << '\n'

int main ( void ) {
int i = 10;
int& ri = i;
SHOW( typeof(i) );
SHOW( typeof(ri) );
SHOW( ( &typeid( ri ) == &typeid(int) ) );
SHOW( typeof( int_ref() ) );
SHOW( ( &typeid( int_ref() ) == &typeid(int) ) );
}

Output:
typeof(i) = int
typeof(ri) = int
( &typeid( ri ) == &typeid(int) ) = 1
typeof( int_ref() ) = int
( &typeid( int_ref() ) == &typeid(int) ) = 1


It looks as though ri has type int for all practical purposes. Also,
whatever is returned by int_ref() masquerades really well as an int.


Best

Kai-Uwe Bux

[Roland Pibinger]

Re: Is &quot;reference &quot; a (different?) type?

On Sun, 15 Jul 2007 06:08:27 +0200, Kai-Uwe Bux wrote:

>int main ( void ) {
int i = 10;
int& ri = i;
SHOW( typeof(i) );
SHOW( typeof(ri) );
SHOW( ( &typeid( ri ) == &typeid(int) ) );
SHOW( typeof( int_ref() ) );
SHOW( ( &typeid( int_ref() ) == &typeid(int) ) );
>}
>
>Output:
>typeof(i) = int
>typeof(ri) = int
>( &typeid( ri ) == &typeid(int) ) = 1
>typeof( int_ref() ) = int
>( &typeid( int_ref() ) == &typeid(int) ) = 1
>
>It looks as though ri has type int for all practical purposes. Also,
>whatever is returned by int_ref() masquerades really well as an int.

Interesting. After initialization a reference is merely an 'alias',
i.e. an 'alternate name', for the referenced object
(http://www.parashift.com/c++-faq-lite/references.html). The type of
the reference becomes indistinguishab le from the type of the
referenced object after initialization.


--
Roland Pibinger
"The best software is simple, elegant, and full of drama" - Grady Booch