regarding FAQ 8.5

[jammie_linux]

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" , the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
cout << ref << endl;
return 0;
}

I used g++ to compile the above program.

[Mike Wahler]

<ja**********@yahoo.com> wrote in message
news:11**********************@g43g2000cwa.googlegr oups.com...

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" , the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?
Because it's valid.

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
This is equivalent to

i = j;

The reference still refers to 'i' (iow 'ref' is still
an alias for 'i').

cout << ref << endl;
This should print 0. Does it?
return 0;
}

I used g++ to compile the above program.

-Mike

[Bob Hairgrove]

On 11 Oct 2005 11:47:25 -0700, ja**********@yahoo.com wrote:

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" ...
You cannot.
... the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
// OK: ref is an alias for i.
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
You assign the value of j to i here. You haven't "changed the
reference variable", nor can you.
cout << ref << endl;
return 0;
}

I used g++ to compile the above program.

--
Bob Hairgrove
No**********@Home.com

[Martin Vorbrodt]

<ja**********@yahoo.com> wrote in message
news:11**********************@g43g2000cwa.googlegr oups.com...

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" , the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
cout << ref << endl;
return 0;
}

I used g++ to compile the above program.

No, you're not.
In this case ref = j; is equivalent to i = j;
cout << i << endl; after you do ref=j; you'll see for yourself.

Martin

[Greg Comeau]

In article <11**********************@g43g2000cwa.googlegroups .com>,
<ja**********@yahoo.com> wrote:

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" , the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
cout << ref << endl;
return 0;
}

I used g++ to compile the above program.

That does not "reseat" ref. One way to look at this is to
consider that ref becomes another name for i. Therefore,
since you can say i = j, which doesn't mean i become j
but that j's value is copies into i, the same applies here.
Since ref is another name for i, then ref = j is as if it
were i = j.

If you prefer to stay with the constand pointer analogy you
might have:

int * const ref = &i;

i = j; // as usual
*ref = j; //since *ref is also another 'name' for i, then as if: i = j

ref = &j // error
--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
World Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.
Comeau C/C++ with Dinkumware's Libraries... Have you tried it?

[Greg Comeau]

In article <di**********@panix2.panix.com>,
Greg Comeau <co****@comeaucomputing.com> wrote:

In article <11**********************@g43g2000cwa.googlegroups .com>,
<ja**********@yahoo.com> wrote:

Hi,
According to the FAQ [8.5] "How can you reseat a reference to make it
refer to a different object?" , the reference is like a constant
pointer.Then why the following program is not giving any compilation
error/s ?

#include <iostream>
using namespace std;

int main(void)
{
int i=9,j=0;
int &ref=i;
cout<< ref << endl;
ref= j; // I'm changing the reference variable.
cout << ref << endl;
return 0;
}

I used g++ to compile the above program.

That does not "reseat" ref. One way to look at this is to
consider that ref becomes another name for i. Therefore,
since you can say i = j, which doesn't mean i become j
but that j's value is copies into i, the same applies here.
Since ref is another name for i, then ref = j is as if it
were i = j.

If you prefer to stay with the constand pointer analogy you
might have:

int * const ref = &i;

i = j; // as usual
*ref = j; //since *ref is also another 'name' for i, then as if: i = j

ref = &j // error

Also, remember that most operations on references are
actually operations on the thing it references. IOWs,
in this case, after ref is "bound", operations on ref
are actually operations on i.
--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
World Class Compilers: Breathtaking C++, Amazing C99, Fabulous C90.
Comeau C/C++ with Dinkumware's Libraries... Have you tried it?