How to free an array of structures inside a structure?

Question

I have the following code.:

struct O1 {
    int a;
    int b;
    int c;
};

struct O2 {
    int f;
    int g;
    struct O1 *array;      //pointer to an array of structs O1
};

struct O2 *list = malloc(25 * sizeof(struct O1));        //array of 25 structs O2

struct O1 *innerlist = malloc(sizeof(struct O1));    //array of structs O1 contained inside each struct O2

innerlist[0].a = 1;
innerlist[0].b = 2;
innerlist[0].c = 3;

(*list)[0].f = 1;
(*list)[0].g = 1;
(*list)[0].array = &innerlist[0];    //filling the parameters via some loop for example

So my question is, did I allocate the arrays correctly?

For the inner structure contained inside each struct O2, I only wanted an array with just one element (one struct O1) that I could then expand further via realloc for example but I tried storing more struct O1s in struct O2 and it works so there must have been more memory allocated somewhere even though I did not multiply the sizeof by anything when creating the array of struct O1.

I also have a question, how do I free the arrays?

I guess I must first free the inner array first but I am having trouble here.

I have created a function for freeing the memory that takes the array of struct O2 as a parameter.

But when I type free(list->obj); it seems to be freeing the obj of the first struct O2 in the list and free(list[n]->obj) does not seem to be working.

Numeric identifiers like struct 01 will not compile. Did you intend capital-O struct O1? There are some missing ; too after the struct definitions. Is this your code? — Weather Vane
– Weather Vane, Commented Dec 3, 2015 at 17:47
This code does not compile. Please show us some compilable code that exhibits the issue. — dbush
– dbush, Commented Dec 3, 2015 at 17:52
Also (*list)[0] is syntactically equivalent to *((*list)+0). This is wrong I think. You said your example compiles. Can you post that example please? — Ely
– Ely, Commented Dec 3, 2015 at 17:54
It's difficult to answer when you provide that has too many compiler errors. — R Sahu
– R Sahu, Commented Dec 3, 2015 at 17:56

001 · Accepted Answer · 2015-12-03 17:54:59Z

2

1 Create an array of O2:

struct O2 *list = malloc(25 * sizeof(struct O2));

2 Allocate space for 1 O1 in each O2

for (int i = 0; i < 25; i++) {
    list[i].array = malloc(sizeof(struct O1));
}

3 Free the mem

for (int i = 0; i < 25; i++) {
    free(list[i].array);
}
free(list);

answered Dec 3, 2015 at 17:54

001

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9 Comments

Daeto Over a year ago

This looks good. Will that allocate an array of structs 01 though but with just one element? I mean, can I expand it later and add more structs 01 to the array?

001 Over a year ago

Yes. In the original post you said you wanted just one element. You can allocate more - malloc(sizeof(struct O1) * 100); - or realloc later.

Daeto Over a year ago

Yes thanks a lot. How about filling the elements of struct 01 though? Lets say inside a function where the argument is the array of structs 02. I have tried (*list)[i].obj->id = 1; for example but this does not work.

001 Over a year ago

You would treat it like a regular array: list[x].array[y].a = 1;

Daeto Over a year ago

Something is wrong with my allocation as the program crashes. My function takes a parameter struct 02 **ptr which is a pointer to a pointer to the array. I then allocate the main array like this *ptr = malloc(25 * sizeof(struct O2)); Then run the for loop like this for (int i = 0; i < 25; i++) { ptr[i]->array = malloc(sizeof(struct O1)); And try to fill the parameters of the structures like this: (*ptr)[n].f = 1; (*ptr)[n].array[0].a = 1;

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