array of pointers in struct

**cs32** · 04-06-2008

Here's a small piece of code that I'm having some trouble with.

Code:

typedef struct my_struct {
    ... 
    int *current_items;
} my_struct;

my_struct *s = malloc(sizeof(my_struct));
s->current_items = malloc(5 * sizeof(int));

I get a compile error stating that assignment makes integer from pointer without a cast. Is this happening because I'm not casting malloc? I've seen comments on this board instructing people not to cast from malloc, so I'm a little bit lost as to what I'm doing wrong here.

Also, when I attempt to compare a pointer to a NULL value in the following manner:

Code:

if (s->current_items[1] == NULL) {
    ....
}

Or this:

Code:

s->current_items[1] = *p;

I also get compile errors. Do I need to cast here, or am I totally off?

Thanks.

**Salem** · 04-06-2008

You get "makes pointer from integer without a cast" from a malloc call, typicially by failing to include stdlib.h

**Elysia** · 04-06-2008

The boards are right in that you shouldn't cast malloc, because it typically hides the warning that you're doing an implicit call to malloc which is very bad.

**ssharish2005** · 04-06-2008

Code:

if (s->current_items[1] == NULL) {
    ....
}

Well, yes this would give compiler warning. Since you are comparing a NULL pointer with the integer. And the second example as well.

Code:

s->current_items[1] = *p;

You are trying to assign a pointer to a integer array element. Which is wrong again unless the array was declared as an array of integer pointers.

ssharish

**cs32** · 04-09-2008

OK - I'm a little bit confused. Why do I get a compiler warning with the following code:

Code:

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

int
main(int argc, char *argv[]) {
  
  int foo[5];
  int *i = malloc(sizeof(int));

  foo[0] = (int) i;
  printf("%d\n", foo[0]);

  return 0;
}

warning: assignment makes integer from pointer without a cast

When I explicitly cast the pointer to an integer, the warning goes away. Isn't the pointer supposed to be an integer? Why does this happen?

**Elysia** · 04-09-2008

No, a pointer is not an integer. It's a pointer, and a pointer is a variable that contains an address. Do not cast anything unless you know what you're doing.
You need to dereference the pointer to get the value at the address:
foo[0] = *i;
http://cpwiki.sf.net/A_pointer_on_pointers

**laserlight** · 04-09-2008

Isn't the pointer supposed to be an integer?

No, a pointer is not an integer. It does not make sense to cast a pointer to int. What are you trying to do?

Note that you did not free() what you malloc()ed.

**cs32** · 04-09-2008

Originally Posted by laserlight

No, a pointer is not an integer. It does not make sense to cast a pointer to int. What are you trying to do?

I'm trying to create an array of pointers. I'm obviously way off base here.

**Elysia** · 04-09-2008

But that's easy.

Code:

int* foo[5];

A pointer is a type and you can create an array of any type (except void).

**dwks** · 04-09-2008

Notice all of the malloc calls, Elysia? . . . .

Maybe you want something like this.

Code:

int **data = malloc(N * sizeof(*data));

Now data[0] through data[N-1] are valid pointers. If you want to make each of those pointers point to dynamically allocated memory as well, you could use something like this:

Code:

int **data = malloc(N * sizeof(*data));
size_t x;

for(x = 0; x < N; x ++) {
    data[x] = malloc(sizeof(**data));
}

Note that sizeof(*data) is just like sizeof(int *), because data is an int **; and sizeof(**data) is like sizeof(int). It's a bit easier to change the type of a variable if you use it in the sizeof, however, instead of the type itself directly.

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