Thread: Incompatible pointer type !?

Post a small "working" (in your case, "broken") example that you're attempting to compile.


Quzah.

> head->ptr2struct[0] = obj; and head->ptr2struct[1] = obj;
You don't have an array of pointers.

head->ptr2struct = obj;
Would be valid in this instance.

Originally Posted by nilathinesh

since an array and pointer are interchangeble

Common myth. This applies only to a few situations, and you're doing it wrong, anyway.

Code:

head->ptr2struct[0] = obj;
head->ptr2struct[1] = obj;

ptr2struct is of type struct whatevr*. obj is too. But ptr2struct[0] is of type struct whatevr, not a pointer to it.

Code:

struct structag
{
datatype datavariable;
struct structag *ptr2struct;     ===> this is * equivalently, int ptr2struct[1]; can also be used is it !?
};

No, they're not the same. The pointer places a pointer in the struct. The array of size 1 places an array of size 1 in the struct (which is impossible, because you'd need the complete definition of the struct at this point, which you don't have; and anyway, it would be like placing a simple instance of the struct in the struct, i.e. infinite size.)

so i tried like allocating the memory at runtime one by one and decided to add like..array accessing

But getting this incompatible error...

You're allocating objects and then you're trying to assign pointers to those objects.

One of my collegue saying that it will be compatible in C99.. but i dont know.. since i have not tried, it .. any info.. on this... plz..

No, it will not. You've got a simple type mismatch, and never in a thousand years will this ever be compatible.
The C99 feature is for undefined size arrays, but it works differently, and definitely not for nesting itself.

You want something like this then

Code:

struct structag
{
    datatype datavariable;
    struct structag **ptr2struct;
};

head->ptr2struct = malloc ( 2 * sizeof *head->ptr2struct );
head->ptr2struct[0] = obj;
head->ptr2struct[1] = obj;

Change 2 for some other number, if you've stored however many you want somewhere else.

Why do you keep trying to index these pointers as if they were arrays?
Why do you keep casting malloc?
Why aren't you paying attention to anything anyone tells you?


Quzah.

So what you're really after is a linked list.

Code:

#include <stdio.h>
struct structag
{
  int a;
  struct structag *ptr;
};
int main()
{
struct structag *head, *current, *obj;

current = malloc( sizeof *current );
head = current;

current->a =10;
obj = malloc(sizeof *obj );
current->ptr = obj;
current = obj;

current->a =20;
obj = malloc(sizeof *obj );
current->ptr = obj;
current = obj;

current->ptr = NULL; /* end of list */

}

Now wrap those up into concepts such as
- initialise list
- append to like
- create node

Why you shouldn't cast malloc.

Consider:

Code:

char *ptr = "abcde";
char c;

c = ptr[ 0 ];

This is fine. Indexing further down the line is also fine, so long as you actually have something valid to point to. You were doing this:

Code:

struct foo *ptr = malloc( sizeof *ptr );
...
ptr[ 1 ].something = somethingelse;

You can't index 1, because you've only allocated enough space for one structure. Thus, the only valid index above would be:

Code:

ptr[ 0 ].something = somethingelse;

Or, if you prefer:

Code:

ptr->something = somethingelse;

You were trying to walk off the end of your allocated space. That's why it wasn't working.

Yes, any single pointer can be done like so:

Code:

int *ptr;
int x;

ptr = &x;

ptr[0] = 10;

It's fine to use array style indexing, so long as you're actually accessing valid memory.

Code:

ptr[10] = foo;

You can't just do that if you don't actually have something valid at that location.


Quzah.

> but still i could not ensure is it compatible with C99
By making it compatible with C89 of course.