what is wrong with this

[free2cric]

Hi,
I have declared pointer of pointers **a;
so In a loop I assign a block to a pointer and put a value in it
and then I want to print these values.

My following program doesnt work.

Also How to do same program with int *a[]; (array of pointers).
Any help would really help.

#include <stdio.h>
#include <conio.h>
#include <malloc.h>

void main()
{
int **a;
int v=0;

for (v=0;v<5;v++)
{
*(a+v)=(int *)malloc(sizeof (int));
**(a+v)=v;
}

for (v=0;v<5;v++)
{
printf("%d\n",* *(a+v));
}

exit(0);

}

[A. Sinan Unur]

fr*******@yahoo .com wrote in news:1140783951 .891486.192110
@t39g2000cwt.go oglegroups.com:

Any help would really help.
OK, then.
#include <stdio.h>
#include <conio.h>
conio.h is a non-standard header. Don't see any need for any nonstandard
constructs below, so omit this.
#include <malloc.h>
malloc.h is a non-standard header. You should:

#include <stdlib.h>

for malloc.
void main()
This is a non-standard prototype for main. In this case, you should use:

int main(void)
{
int **a;
int v=0;

for (v=0;v<5;v++)
{
*(a+v)=(int *)malloc(sizeof (int));
Don't cast the return value of malloc. It can hide failure to include
stdlib.h.
**(a+v)=v;
BAM! a itself points to nowhere. It should point to a chunk of memory
large enough to hold exactly five pointers to int.
}

for (v=0;v<5;v++)
{
printf("%d\n",* *(a+v));
}

exit(0);

return 0;
}

is more conventional.

Sinan
--
--
A. Sinan Unur <1u**@llenroc.u de.invalid>
(reverse each component and remove .invalid for email address)

[pranav.choudhary]

Perhaps i know where you have got this wrong. You declared
int **a;
can you read the above declaration, what does it say?? It says "a is a
pointer to pointer to int". So a is large enough to hold one pointer to
integer. But in your code you assume it large enough to hold 5 pointers
to integer, ie. a is not an array of 5 pointers to int, but just a
(single)pointer to pointer to int. To make your program work we have to
say

int **a;
a = (int **)malloc(5 *sizeof(int *));

I hope things are clear now.

[A. Sinan Unur]

"pr************ **@gmail.com" <pr************ **@gmail.com> wrote in
news:11******** **************@ i40g2000cwc.goo glegroups.com:

Perhaps i know where you have got this wrong.
Who got what wrong? See <URL:http://cfaj.freeshell. org/google/>
int **a;
a = (int **)malloc(5 *sizeof(int *));
#include <stdlib.h>

int **a = malloc(5 * sizeof(*a));

if ( a ) {

}

Casting the return value of malloc can hide errors due to failure to
include stdlib.h.

By using the form above, the malloc call remains the same even if you
change the type of a.
I hope things are clear now.

Not really.

Sinan

--
A. Sinan Unur <1u**@llenroc.u de.invalid>
(reverse each component and remove .invalid for email address)

[pranav.choudhary]

I agree with that. My mail was not in reply to yours but to the
original mail. There was something else that was wrong with the posted
code and not the issue of not being able to include a header file. The
guy was simply treatin int **a; as int *a[5];

[Richard Bos]

"pr************ **@gmail.com" <pr************ **@gmail.com> wrote:

I agree with that.
With _what_? Learn to tell Google Broken Beta to post with context, or
get a newsreader.
My mail was not in reply to yours but to the original mail.

There are no mails. There are only posts. Usenet is not an e-mail
service. Usenet is not the property of Google "do evil" Inc. Usenet is
an international network of newsgroups, with their own netiquette, far
preceeding the existence of Google. Learn to use it properly, please.

Richard

[Richard G. Riley]

On 2006-02-24, fr*******@yahoo .com <fr*******@yaho o.com> wrote:

Hi,
I have declared pointer of pointers **a;
so In a loop I assign a block to a pointer and put a value in it
and then I want to print these values.

My following program doesnt work.

Also How to do same program with int *a[]; (array of pointers).
Any help would really help.

#include <stdio.h>
#include <conio.h>
#include <malloc.h>

void main()
{
int **a;
int v=0;

for (v=0;v<5;v++)
{
*(a+v)=(int *)malloc(sizeof (int));
You have not yet allocated an area of memory to store your int
pointers. a is pointing to nothing valid.

There are other horrible things there too but if you get the
first bit right, the rest should logically follow as you get to grips
with pointers and pointers to pointers. Good luck!
**(a+v)=v;
}

for (v=0;v<5;v++)
{
printf("%d\n",* *(a+v));
}

exit(0);

}

--
Remove evomer to reply

[CBFalconer]

fr*******@yahoo .com wrote:

I have declared pointer of pointers **a;
so In a loop I assign a block to a pointer and put a value in it
and then I want to print these values.

My following program doesnt work.

Also How to do same program with int *a[]; (array of pointers).
Any help would really help.

#include <stdio.h>
#include <conio.h>
no such standard include file. Remove this.
#include <malloc.h>
No such standard include file. You probably want stdlib.h

void main()
All bets are off. main returns int. Say so.
{
int **a;
int v=0;

for (v=0;v<5;v++)
The use of blanks is allowed, and adds legibility. No extra charge
is made for them.

for (v = 0; v < 5; v++)
etc.
{
*(a+v)=(int *)malloc(sizeof (int));
Do not cast the return value of malloc. It only hides errors.
Besides, how can the left side, which has no associated memory,
hold the result? a has been defined to hold one single pointer to
a pointer to an int, and has never been initialized. Who knows
what (a + v) refers to. Even if the program somehow survived the
earlier gross errors, this made it go KA-BOOM. The approved way to
use malloc is:

ptr = malloc(SOMENUMB ER * sizeof *ptr);

followed by checking for success, i.e. (ptr != NULL). Note that
the *s above are not the same thing, one is a multiplicative
operator, and one is a dereferencing operator.

In your case you would need multiple mallocs, something like:

if (a = malloc(5 * sizeof *a)) {
for (n = 0; n < 5; n++) {
if (*(a + n) = malloc(sizeof *(a + n))) {
/* now you have a place to store an int */
}
else handlemallocfai lure();
}
else handlemallocfai lure();

and somewhere there should be a define for the magic number 5.
**(a+v)=v;
}
for (v=0;v<5;v++)
{
printf("%d\n",* *(a+v));
}
exit(0);
}

If you want to post a followup via groups.google.c om, don't use the
broken "Reply" link at the bottom of the article. Click on "show
options" at the top of the article, then click on the "Reply" at
the bottom of the article headers.

More details at: <http://cfaj.freeshell. org/google/>
Also see <http://www.safalra.com/special/googlegroupsrep ly/>

--
Some informative links:
news:news.annou nce.newusers
http://www.geocities.com/nnqweb/
http://www.catb.org/~esr/faqs/smart-questions.html
http://www.caliburn.nl/topposting.html
http://www.netmeister.org/news/learn2quote.html

[Keith Thompson]

rl*@hoekstra-uitgeverij.nl (Richard Bos) writes:
[...]

There are no mails. There are only posts. Usenet is not an e-mail
service. Usenet is not the property of Google "do evil" Inc. Usenet is
an international network of newsgroups, with their own netiquette, far
preceeding the existence of Google. Learn to use it properly, please.

.... by reading <http://cfaj.freeshell. org/google/>.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[free2cric]

ok guys whaterver u told , it helped me..
but who will free the allocated memory.. ur fathers????
Keith Thompson wrote:

rl*@hoekstra-uitgeverij.nl (Richard Bos) writes:
[...]

There are no mails. There are only posts. Usenet is not an e-mail
service. Usenet is not the property of Google "do evil" Inc. Usenet is
an international network of newsgroups, with their own netiquette, far
preceeding the existence of Google. Learn to use it properly, please.

... by reading <http://cfaj.freeshell. org/google/>.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.