returning function pointer

[Pradyut Bhattacharya]

Hi,
in this program i cannot retrive the values from the pointer j....
The code :-

--------------------------------------------------------------------------------------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()
{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

*j = func2();
printf("\nTraversing second list\n");

printf("\n%d", *j);

}

void func(int a[], int size)
{
int i=0;
for (i=0; i<=size; i++)
printf("%d\n", a[i]);
}

int*func2()
{

int b[] = {112,3,4,5,6,7,9,4,10};

return *b;
}

-------------------------------------------------------------------------------------------------------------

Any help

Thanks
Pradyut
http://pradyut.tk
India

[Default User]

Pradyut Bhattacharya wrote:

Hi,
in this program i cannot retrive the values from the pointer j....
The code :-

----------------------------------------------------------------------
----------------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()

main() returns int ALWAYS.

{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

The number of elements in your array is three. Why pass two?

>
*j = func2();
printf("\nTraversing second list\n");

printf("\n%d", *j);

You're obviously not traversing, but accessing the first element.

>
}

void func(int a[], int size)
{
int i=0;
for (i=0; i<=size; i++)

Change that to i < size

printf("%d\n", a[i]);
}

int*func2()
{

int b[] = {112,3,4,5,6,7,9,4,10};

return *b;

Yikes. You pass back an int from a function that's supposed to be
returning an int *. Presumbably you meant to type:

return b;

However, that's not right either. b is a local array, you can NOT use
it outside of its scope, which happens to be the function func2().

You'll either have to dynamically allocate an array and populate it, or
make b static.

Any help

Turn up the warning level on your compiler.


Brian

[pete]

Default User wrote:

>
Pradyut Bhattacharya wrote:

Hi,
in this program i cannot retrive the values from the pointer j....
The code :-

----------------------------------------------------------------------
----------------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()

main() returns int ALWAYS.

{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

The number of elements in your array is three. Why pass two?

*j = func2();
printf("\nTraversing second list\n");

printf("\n%d", *j);

You're obviously not traversing, but accessing the first element.

}

void func(int a[], int size)
{
int i=0;
for (i=0; i<=size; i++)

Change that to i < size

printf("%d\n", a[i]);
}

int*func2()
{

int b[] = {112,3,4,5,6,7,9,4,10};

return *b;

Yikes. You pass back an int from a function that's supposed to be
returning an int *. Presumbably you meant to type:

return b;

However, that's not right either. b is a local array, you can NOT use
it outside of its scope, which happens to be the function func2().

You'll either have to dynamically allocate an array
and populate it, or make b static.

Any help

Turn up the warning level on your compiler.

/* BEGIN new.c */

#include <stdio.h>

#define NINE 9

void func(int a[], int size);
int *func2(void);

int main(void)
{
int a[] = {3,2,4};
int *(*j)(void) = func2;

func(a, sizeof a / sizeof *a);
printf("\nTraversing second list\n");
func(j(), NINE);
return 0;
}

void func(int a[], int size)
{
int i;

for (i = 0; size i; i++) {
printf("%d\n", a[i]);
}
}

int *func2(void)
{
static int b[NINE] = {112,3,4,5,6,7,9,4,10};

return b;
}

/* END new.c */
--
pete

[Default User]

Default User wrote:

Pradyut Bhattacharya wrote:

Hi,
in this program i cannot retrive the values from the pointer
j.... The code :-

--------------------------------------------------------------------
-- ----------------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()

main() returns int ALWAYS.

{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

The number of elements in your array is three. Why pass two?

*j = func2();

printf("\nTraversing second list\n");

printf("\n%d", *j);

Oops, I just noticed that you are printing an int pointer as if it were
an int. Yet another mistake. j is an array of int pointers. Possibly
you wanted it to be just an int pointer, which would make the printf ok.


Brian

[Default User]

pete wrote:

void func(int a[], int size);
int *func2(void);

int main(void)
{
int a[] = {3,2,4};
int *(*j)(void) = func2;

func(a, sizeof a / sizeof *a);
printf("\nTraversing second list\n");
func(j(), NINE);
return 0;
}

Hmmm, points for working part of the subject into the solution, but the
OP said RETURNING function pointer.


Brian

[pete]

Default User wrote:

>
pete wrote:

void func(int a[], int size);
int *func2(void);

int main(void)
{
int a[] = {3,2,4};
int *(*j)(void) = func2;

func(a, sizeof a / sizeof *a);
printf("\nTraversing second list\n");
func(j(), NINE);
return 0;
}

Hmmm, points for working part of the subject into the solution,
but the OP said RETURNING function pointer.

I think I did better than that.

OP said "cannot retrive the values from the pointer j"
I made j into a function pointer.
I used the return value from a function call made through j
to access nine values.
And that's how I got OP's original line of code:
printf("\nTraversing second list\n");
to say what it means.

--
pete

[Mark McIntyre]

On Wed, 27 Jun 2007 02:17:16 +0530, in comp.lang.c , "Pradyut
Bhattacharya" <pr******@gmail.comwrote:

>int*func2();

funct returns a pointer to an int...

int*j[8];

j is an array of pointers to ints. I don't think you wanted that...

*j = func2();

you set the first pointer in j to whatever func2 returns.

int b[] = {112,3,4,5,6,7,9,4,10};

b is a local array which is discarded when func2 returns.

return *b;

*b is the value of the first element ie 112,

But your function is expecting a /pointer/ to an int. You should
hopefully get a warning here. If not, turn up warning levels. You have
set j[0] to point to the 112th byte of your computer's memory, which
is probably protected ROM space or something...
I presume you want return an array from a function. You can't do that
(easily). Pass the array as an argument.
--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan

[Default User]

pete wrote:

Default User wrote:

pete wrote:

void func(int a[], int size);
int *func2(void);
>
int main(void)
{
int a[] = {3,2,4};
int *(*j)(void) = func2;
>
func(a, sizeof a / sizeof *a);
printf("\nTraversing second list\n");
func(j(), NINE);
return 0;
}

Hmmm, points for working part of the subject into the solution,
but the OP said RETURNING function pointer.

I think I did better than that.

OP said "cannot retrive the values from the pointer j"
I made j into a function pointer.
I used the return value from a function call made through j
to access nine values.
And that's how I got OP's original line of code:
printf("\nTraversing second list\n");
to say what it means.

Ok, that's not bad.

Brian

[Martin Ambuhl]

Pradyut Bhattacharya wrote:

Hi,
in this program i cannot retrive the values from the pointer j....

We won't bother discussing absurdities of

void main()

If you don't know better by now, since you have done as any civilized
person would and followed the newsgroup and checked the FAQ before
posting, you never will.

The answer to your stated problem is simple. There is no "the pointer j".
You have declared j with

int*j[8];

So j is an array[8] of pointers to int.

[!truth]

On 6 27 , 4 47 , "Pradyut Bhattacharya" <[e-mail]:

Hi,
in this program i cannot retrive the values from the pointer j....
The code :-

---------------------------------------------------------------------------*-----------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()
{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

*j = func2();
printf("\nTraversing second list\n");

printf("\n%d", *j);

}

void func(int a[], int size)
{
int i=0;
for (i=0; i<=size; i++)
printf("%d\n", a[i]);

}

int*func2()
{

int b[] = {112,3,4,5,6,7,9,4,10};

return *b;

}

---------------------------------------------------------------------------*----------------------------------

Any help

Thanks
Pradyuthttp://pradyut.tk
India

It not wise to return a pointer which points to the static.
But if you still want to do so, try to remove the line:
printf("\nTraversing second list\n");

[J. J. Farrell]

On Jun 26, 9:36 pm, !truth <[e-mail]:

On 6 27 , 4 47 , "Pradyut Bhattacharya" in this program i cannot retrive the values from the pointer j....

The code :-

---------------------------------------------------------------------------**-----------------------------------

#include <stdio.h>

void func(int [], int);
int*func2();

void main()
{
int a[] = {3,2,4};

int*j[8];
int i;

func(a, 2);

*j = func2();
printf("\nTraversing second list\n");

printf("\n%d", *j);

}

void func(int a[], int size)
{
int i=0;
for (i=0; i<=size; i++)
printf("%d\n", a[i]);

}

int*func2()
{

int b[] = {112,3,4,5,6,7,9,4,10};

return *b;

}

---------------------------------------------------------------------------**----------------------------------

Any help

It not wise to return a pointer which points to the static.

Whyever not? What does this have to do with the question anyway?

But if you still want to do so, try to remove the line:
printf("\nTraversing second list\n");

Why? How is that relevant to anything?