Use of member access operator on a pointer

Question

I haven't used C in a while, and now i'm trying to go back to it.

My problem is that my code actually works, although I was sure I will get syntax errors. He'res What I want to have: A dynamic array of polygons, each has a dynamic array of points.

struct point{
  int x,y;
};

struct polygon{
  int quantity;
  struct point* point_list;
};

I will then initialize an array of polygons:

struct polygon* poly_array = (struct polygon*) malloc(sizeof(struct polygon)*num);

and also, initialize each polygon's point array:

poly_array[i].quantity = points;
poly_array[i].point_list = (struct point*) malloc (sizeof( struct point) *  poly_array[i].quantity);

Now, what I thought I was doing, was create arrays of pointers to the object. So, to access inner fields, I would need to use the "->" operator. But no, it works with "direct" access to the fields (short version, without the loop code):

poly_array[i].point_list[j].x = i;
poly_array[i].point_list[j].y = j;

and also to print:

printf ("poly %d: (%d, %d)\n", j, poly_array[i].point_list[j].x, poly_array[i].point_list[j].y);

So, to make my question clear, I will ask again: since I'm iterating on pointers to objects, shouldn't I use the -> operator?

You are not creating arrays of pointers and not iterating on pointers to objects. Perhaps you meant to, but that's not what actually happens in your code. — n. m. could be an AI, Mar 08 '16 at 19:44
`type->field` is simply an equivilant to `(*type).field`. It is used so you dont have to explicitly dereference a pointer to get to a type inside. — element11, Mar 08 '16 at 19:46
No you would use polygon ->point_list[i]... the left hand side has to be a pointer , hence the use of arrow symbol to remind you ! — Arif Burhan, Mar 08 '16 at 19:46
Use a `struct` with a flexible array member. This avoids one indirection and seperate allocation. — too honest for this site, Mar 08 '16 at 20:45

score 4 · Accepted Answer · edited May 23 '17 at 12:23

4

In your code, poly_array is of type struct polygon*, so, poly_array[i] will be of type struct polygon. Thus, you have to use the . operator.

same logic applies for point_list[j] also.

To elaborate, quoting C11, chapter §6.5.2.1, Array subscripting (emphasis mine)

syntax

postfix-expression [ expression ]

Constraints

One of the expressions shall have type ‘‘pointer to complete object type’’, the other expression shall have integer type, and the result has type ‘‘type’’.

That said, please see this discussion on why not to cast the return value of malloc() and family in C..

edited May 23 '17 at 12:23

Community

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answered Mar 08 '16 at 19:43

Sourav Ghosh

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so what i have is simply an array that holds actual polygons, yes? – yoad w Mar 08 '16 at 19:49
@yoadw Not really, pointers are not arrays. Concentrate on the data type itself. – Sourav Ghosh Mar 08 '16 at 19:51
so, if I wanted to have an array of pointers to polygons, how would I need to change my code? (I hope it's not too much to ask, thanks!) – yoad w Mar 08 '16 at 20:01
@yoadw array of pointers to polygons? It should be something like `struct polygon* poly_array[10]` and the `malloc()` will also change, but are you sure that's what you need? (10 is arbitrary anyway...) – Sourav Ghosh Mar 08 '16 at 20:03
i don't need it for any specific reason, i'm doing it to refresh memory\knowledge regarding C. this is actually a part of question from some old college test, and i'm messing around with the results. thanks. – yoad w Mar 08 '16 at 20:08
@yoadw You're welcome. You can also [consider accepting an answer that helped you](http://meta.stackexchange.com/q/5234/244062). – Sourav Ghosh Mar 08 '16 at 20:15

score 0 · Answer 2 · answered Mar 08 '16 at 20:42

Short answer is referring to the array by index already dereferences the indexed item, so you are correct using dot syntax.

poly_array[i].point_list[j].x

You could as well use the "->" operator to dereference the item directly without using index operator []. In that case you would use pointer arithmetic.

(((poly_array+i)->point_list)+j)->x

Use of member access operator on a pointer

2 Answers2