how the following two code parts same

[saipathakota]

how the following two code parts same:

#include <stdio.h>

void main ()
{

char i = NULL;

char &q = i;
^^^^

printf ( "%d", i);

printf ("%d", q);
}

O/p.................. 0 0

#include <stdio.h>

void main ()
{

char i = NULL;

char q = i;
^^

printf ( "%d", i);

printf ("%d", q);
}
O/p ................ 0 0

[fred.l.kleinschmidt]

On Dec 7, 1:00 pm, [e-mail] wrote:

how the following two code parts same:

#include <stdio.h>

void main ()
{

char i = NULL;

char &q = i;
^^^^

printf ( "%d", i);

printf ("%d", q);

}

O/p.................. 0 0

#include <stdio.h>

void main ()
{

char i = NULL;

char q = i;
^^

printf ( "%d", i);

printf ("%d", q);

}

O/p ................ 0 0

They are the same except for the lines

char &q = i;

and

char q = i;

--
Fred

[saipathakota]

On Dec 8, 2:03 am, [e-mail] wrote:

On Dec 7, 1:00 pm, [e-mail] wrote:

how the following two code parts same:

#include <stdio.h>

void main ()
{

char i = NULL;

char &q = i;
^^^^

printf ( "%d", i);

printf ("%d", q);

}

O/p.................. 0 0

#include <stdio.h>

void main ()
{

char i = NULL;

char q = i;
^^

printf ( "%d", i);

printf ("%d", q);

}

O/p ................ 0 0

They are the same except for the lines

char &q = i;

and

char q = i;

--
Fred

i am asking how both code is showing same output.

[Lew Pitcher]

sa**********@gmail.com wrote:

how the following two code parts same:

#include <stdio.h>

void main ()

Nope. In a hosted environment, main() returns int by definition. You really mean
int main(void)
here

{

char i = NULL;

You should see a diagnostic indicating that you are converting a pointer to an
integer (type) without a cast. Having said that, i should now be initialized
to zero.

>
char &q = i;

Syntax error. Your program should not compile.

>
printf ( "%d", i);

printf ("%d", q);
}

O/p.................. 0 0

I don't believe you.

>
#include <stdio.h>

void main ()

Again,
int main(void)

{

char i = NULL;

You should see a diagnostic indicating that you are converting a pointer to an
integer (type) without a cast. Having said that, i should now be initialized
to zero.

char q = i;

q is initialized to the contents of i. Since i is initialized to zero and not
changed before this statement, q should now also be set to zero.

>
printf ( "%d", i);

printf ("%d", q);
}
O/p ................ 0 0

OK

So, what's your problem?

--
Lew Pitcher

Master Codewright & JOAT-in-training | Registered Linux User #112576
http://pitcher.digitalfreehold.ca/ | GPG public key available by request
---------- Slackware - Because I know what I'm doing. ------

[saipathakota]

On Dec 8, 2:13 am, Lew Pitcher <[e-mail]:

[e-mail] wrote:

how the following two code parts same:

#include <stdio.h>

void main ()

Nope. In a hosted environment, main() returns int by definition. You really mean
int main(void)
here

{

char i = NULL;

You should see a diagnostic indicating that you are converting a pointer to an
integer (type) without a cast. Having said that, i should now be initialized
to zero.

char &q = i;

Syntax error. Your program should not compile.

printf ( "%d", i);

printf ("%d", q);
}

O/p.................. 0 0

I don't believe you.

#include <stdio.h>

void main ()

Again,
int main(void)

{

char i = NULL;

You should see a diagnostic indicating that you are converting a pointer to an
integer (type) without a cast. Having said that, i should now be initialized
to zero.

char q = i;

q is initialized to the contents of i. Since i is initialized to zero and not
changed before this statement, q should now also be set to zero.

printf ( "%d", i);

printf ("%d", q);
}

O/p ................ 0 0

OK

So, what's your problem?

--
Lew Pitcher

Master Codewright & JOAT-in-training | Registered Linux User #112576http://pitcher.digitalfreehold.ca/ | GPG public key available by request
---------- Slackware - Because I know what I'm doing. ------

i compiled in the vc++ .. o/p is same .. is it true with the C++

[Default User]

Lew Pitcher wrote:

char i = NULL;

You should see a diagnostic indicating that you are converting a
pointer to an integer (type) without a cast.

You mean you might see a diagnostic. There's an excellent chance that
NULL is not a pointer type at all.


Brian

[Default User]

sa**********@gmail.com wrote:

--
Lew Pitcher

Master Codewright & JOAT-in-training | Registered Linux User

Please trim quotes to the minimum necessary for context. Especially
trim the signatures, the parts after -- (I left part in above as an
example).

i compiled in the vc++ .. o/p is same .. is it true with the C++

VC++ can compile C programs. The first code you showed was NOT a C
program, as this:

char &q = i;

Is not a valid C construct. That looks a lot like a reference in C++.
You should probably be posting in comp.lang.c++.

Brian

[jameskuyper]

Default User wrote:

sa**********@gmail.com wrote:

....

i compiled in the vc++ .. o/p is same .. is it true with the C++

VC++ can compile C programs. The first code you showed was NOT a C
program, as this:

char &q = i;

Is not a valid C construct. That looks a lot like a reference in C++.
You should probably be posting in comp.lang.c++.

When you do post to comp.lang.c++, they will tell you that the results
you got are precisely the results you should have expected to get
(ignoring the defective declarations of main()). Save them some time,
and in your first message to that newsgroup, try to explain why it is
that you are surprised by these results. Then they will be able to
explain to you what it is that you're misunderstanding; because you
are misunderstanding something.

[jameskuyper]

Keith Thompson wrote:

sa**********@gmail.com writes:

....

#include <stdio.h>
void main ()
{
char i = NULL;
char &q = i;
^^^^
printf ( "%d", i);
printf ("%d", q);
}

The above is not C. It appears to be C++. Consulting your C++
textbook or the C++ FAQ for information about "references" should
answer your question. Failing that, try comp.lang.c++.

#include <stdio.h>
void main ()

Wrong. Make this "int main(void)" (or "int main()" if you mean to
write C++).

{
char i = NULL;

NULL is a null *pointer* constant. Assigning it to a char object
might happen to work, but it makes no sense. You probably mean

Given that it is C++ code, it will work. In C++, NULL has to have a
value of 0; it cannot have a pointer type.

[Keith Thompson]

ja*********@verizon.net writes:

Keith Thompson wrote:

>sa**********@gmail.com writes:

[...]

{
char i = NULL;

NULL is a null *pointer* constant. Assigning it to a char object
might happen to work, but it makes no sense. You probably mean

Given that it is C++ code, it will work. In C++, NULL has to have a
value of 0; it cannot have a pointer type.

True, but using NULL for anything other than a pointer value is poor
style in any case.

--
Keith Thompson (The_Other_Keith) <ks***@mib.org>
Looking for software development work in the San Diego area.
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[James Kuyper]

Keith Thompson wrote:

ja*********@verizon.net writes:

>Keith Thompson wrote:

>>sa**********@gmail.com writes:

[...]

>>>{
char i = NULL;
NULL is a null *pointer* constant. Assigning it to a char object
might happen to work, but it makes no sense. You probably mean

Given that it is C++ code, it will work. In C++, NULL has to have a
value of 0; it cannot have a pointer type.

True, but using NULL for anything other than a pointer value is poor
style in any case.

Agreed.