problem with the code

[ramu]

Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);
}
#include<stdio. h>
int main()
{
char *ch;
ch = func();
*ch = 'A';
pritnf("%c",*ch );
}

[ramu]

On Aug 25, 11:22 am, ramu <ramu....@gmail .comwrote:

Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);}

#include<stdio. h>
int main()
{
char *ch;
ch = func();
*ch = 'A';
pritnf("%c",*ch );

}

pritnf("%c",*ch );

>Sorry. It should be printf("%c", *ch);

[fjblurt]

On Aug 24, 11:22 pm, ramu <ramu....@gmail .comwrote:

Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);}

#include<stdio. h>
int main()
{
char *ch;
ch = func();
*ch = 'A';
pritnf("%c",*ch );

}

Looks like homework. How about if you explain whether *you* think
there's a problem, and why or why not. And don't forget, when you
submit the assignment, to acknowledge and cite as references the posts
of those who replied to you.

[Barry Schwarz]

On Sun, 24 Aug 2008 23:22:01 -0700 (PDT), ramu <ra******@gmail .com>
wrote:

>Hi,
Is there any problem with the following code?

Yes. It invokes undefined behavior.

>
Regards

char *func()
{
char c;

c is an automatic variable. It comes into existence at the start of
this block of code.

return (&c);

And it goes out of existence when the function returns.

>}
#include<stdio .h>
int main()
{
char *ch;
ch = func();

At this point ch receives the address of c which has just gone out of
existence. (By definition, the value in ch becomes indeterminate.)

*ch = 'A';

Here you attempt to store the value 'A' into an object that no longer
exists. (Technically, you are trying to evaluate an indeterminate
value which invokes undefined behavior.)

pritnf("%c",*ch );

This statement also invokes undefined behavior by trying to evaluate
the address in ch.

>}

--
Remove del for email

[Default User]

ramu wrote:

Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);
}

This returns a pointer to a local variable. That variable goes out of
scope when the function returns. Any use of the pointer is undefined
behavior.

Brian

[jacob navia]

Barry Schwarz wrote:

On Sun, 24 Aug 2008 23:22:01 -0700 (PDT), ramu <ra******@gmail .com>
wrote:

>Hi,
Is there any problem with the following code?

Yes. It invokes undefined behavior.

>Regards

char *func()
{
char c;

c is an automatic variable. It comes into existence at the start of
this block of code.

> return (&c);

And it goes out of existence when the function returns.

>}
#include<stdio .h>
int main()
{
char *ch;
ch = func();

At this point ch receives the address of c which has just gone out of
existence. (By definition, the value in ch becomes indeterminate.)

> *ch = 'A';

Here you attempt to store the value 'A' into an object that no longer
exists. (Technically, you are trying to evaluate an indeterminate
value which invokes undefined behavior.)

> pritnf("%c",*ch );

This statement also invokes undefined behavior by trying to evaluate
the address in ch.

>}

FINE, now *you* did *his* homework.

Isn't clc great?
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[jacob navia]

Default User wrote:

ramu wrote:

>Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);
}

This returns a pointer to a local variable. That variable goes out of
scope when the function returns. Any use of the pointer is undefined
behavior.

Brian

Why do you do his homework?

clc is there to get
(1): "off topic" remarks for most interesting posts, and
(2) Do the homework of lazy students.

--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[Richard Heathfield]

ramu said:

Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);
}
#include<stdio. h>
int main()
{
char *ch;
ch = func();
*ch = 'A';
pritnf("%c",*ch );
}

Yes, there is a problem with that code (even assuming you meant printf
rather than pritnf).

Perhaps you could explain what you're trying to achieve? That would allow
us to suggest a good way to fix it.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Richard]

Richard Heathfield <rj*@see.sig.in validwrites:

ramu said:

>Hi,
Is there any problem with the following code?

Regards

char *func()
{
char c;
return (&c);
}
#include<stdio .h>
int main()
{
char *ch;
ch = func();
*ch = 'A';
pritnf("%c",*ch );
}

Yes, there is a problem with that code (even assuming you meant printf
rather than pritnf).

Perhaps you could explain what you're trying to achieve? That would allow
us to suggest a good way to fix it.

Why not tell him the returned address of c in func() us meaningless in
c.l.c land (it might be useful if you do not dereference it in some sort
of diagnostic tool)?

Why not tell him that printf() is possibly spelt wrong? Hey maybe he had
a function alled "pritnf" elsewhere?

And we wont even mention the main return code...

[raashid bhatt]

On Aug 24, 11:22*pm, ramu <ramu....@gmail .comwrote:

Hi,
* * * Is there any problem with the following code?

Regards

char *func()
{
* *char c;
* *return (&c);}

#include<stdio. h>
int main()
{
* *char *ch;
* *ch = func();
* **ch = 'A';
* * pritnf("%c",*ch );

}

as u see c is a local stack variable as soon a function returns the
its stack is cleared and it goes out of scope!