dynamic allocation of array of pointers

Question

The following code gives a segmentation fault. I am not able to figure out as to why. Please see..

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int **ptr;
    int *val;
    int x = 7;
    val = &x;
    *ptr = (int *)malloc(10 * sizeof (*val));
    *ptr[0] = *val;
    printf("%d\n", *ptr[0] );

    return 0;
}

on debugging with gdb, it says:

Program received signal SIGSEGV, Segmentation fault.

0x0804843f in main () at temp.c:10

*ptr = (int *)malloc(10 * sizeof (*val));

Any help regarding the matter is appreciated.

I basically want to dynamically allocate an array of pointers. — Vivek Sethi
– Vivek Sethi, Commented Sep 25, 2012 at 9:13
You can't do *ptr until you have first given ptr a value. That's why it segfaults. — ams
– ams, Commented Sep 25, 2012 at 9:23
*ptr = (int *)malloc(10 * sizeof (*val)); is just allocating 10*intsize. Dont know what you are trying to do — auny
– auny, Commented Sep 25, 2012 at 9:23
@rjayavrp I get one in my head when I read the code. It hurts. If you don't get one then you're just lucky. — ams
– ams, Commented Sep 25, 2012 at 9:25

Alok Save · Accepted Answer · 2012-09-25 09:25:06Z

15

int **ptr; 
*ptr = (int *)malloc(10 * sizeof (*val));

First statement declares a double pointer.
Second dereferences the pointer. In order that you are able to dereference it the pointer should point to some valid memory. it does not hence the seg fault.

If you need to allocate enough memory for array of pointers you need:

ptr = malloc(sizeof(int *) * 10);

Now ptr points to a memory big enough to hold 10 pointers to int.
Each of the array elements which itself is a pointer can now be accessed using ptr[i] where,

i < 10

edited Sep 25, 2012 at 9:25

answered Sep 25, 2012 at 9:17

Alok Save

207k56 gold badges448 silver badges539 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Vivek Sethi Over a year ago

But how come it does not show seg fault in some other machine and for me it does?

Lightness Races in Orbit Over a year ago

@VivekSethi When you make mistakes like this you are invoking undefined behaviour. It is not guaranteed that you will get a segmentation fault; that is just the result of dereferencing some invalid memory locations. It is essentially pot luck as to which invalid memory location you are dereferencing since your pointer is uninitialised and holds some arbitrary, unknown underlying value (usually whatever was left there by some other function or program). So you see, it's completely non-deterministic that this will be "caught".

Dimitar Slavchev Over a year ago

When you declare the pointer it receives some value that was in the piece of memory that the computer gave it. There are two options then: 1. If that value is a valid pointer, within the memory your program had allocated, it will not die with Segmentation Fault. You may be writing into other data though. 2. If the value is not within the program's memory, then you are trying to use memory that doesn't belong to your process (or is invalid) hence you get segmentation fault. This is the same a Lightness said above, just with other words.

Kludas · Accepted Answer · 2012-09-25 09:19:22Z

4

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int **ptr;
    int x;

    x = 5;

    ptr = malloc(sizeof(int *) * 10);
    ptr[0] = &x;
    /* etc */

    printf("%d\n", *ptr[0]);

    free(ptr);
    return 0;
}

answered Sep 25, 2012 at 9:19

Kludas

5714 silver badges7 bronze badges

Comments

Whoami · Accepted Answer · 2012-09-25 09:34:18Z

3

See the below program, perhaps, it helps to understand better.

#include<stdio.h>
#include <stdlib.h>
int main(){

/* Single Dimention */

int *sdimen,i;
sdimen = malloc ( 10 * sizeof (int));
/* Access elements like single diminution. */
sdimen[0] = 10;
sdimen[1] = 20;

printf ("\n.. %d... %d ", sdimen[0], sdimen[1]);

/* Two dimention ie: **Array of pointers.**  */

int **twodimen;

twodimen = malloc ( sizeof ( int *) * 10);

for (i=0; i<10; i++) {
  twodimen[i] = malloc (sizeof(int) * 5);

}

/* Access array of pointers */

twodimen[0][0] = 10;
twodimen[0][3] = 30;
twodimen[2][3] = 50;

printf ("\n %d ... %d.... %d ", twodimen[0][0], twodimen[0][3], twodimen[2][3]);
return 0;
}

Hope this helps.. ;).

answered Sep 25, 2012 at 9:34

Whoami

14.5k20 gold badges91 silver badges144 bronze badges

1 Comment

Vivek Sethi Over a year ago

well, this was not exactly what I was looking for. Anyways, it did clear things out in my mind. Thank you.

Karthik G · Accepted Answer · 2013-03-20 07:58:01Z

0

Conceptually if you are using **ptr, then you need to alloacte memory for ptr & *ptr to defrence **ptr.

But in you case you are alloacting memory only for *ptr,if your compiler is smart enough its alloacting memory for ptr(one pointer location) to link *ptr,hence it could able to link ptr->ptr->*ptr.Hence you are not getting Seg Fault.

answered Mar 20, 2013 at 7:58

Karthik G

111 bronze badge

Comments

tanweer alam · Accepted Answer · 2017-06-20 11:19:16Z

0

include

int main()    
{    
    int **ptr;    
    int *val;    
    int x = 7;    
    val = &x;    
    ptr = (int**)malloc(sizeof(int**));    
    *ptr = (int *)malloc(10 * sizeof (*val));    
    *ptr[0] = *val;    
    printf("%d\n", *ptr[0] );    
    return 0;    
}

Above would work. You can find the difference and understand the reason as well.

Bottom line you were de-referencing **ptr without allocating memory to it.

answered Jun 20, 2017 at 11:19

tanweer alam

4374 silver badges6 bronze badges

Collectives™ on Stack Overflow

dynamic allocation of array of pointers

5 Answers 5

3 Comments

Comments

1 Comment

Comments

include

include

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

3 Comments

Comments

1 Comment

Comments

include

include

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related