Question

I currently have the following code in my program:

struct mystruct
{
    int x;
    struct y *list[10];
}

mystruct *h1=(mystruct *)malloc(sizeof(mystruct));

I want to declare the array list dynamically (using malloc()) when I declare the structure. Could anyone tell me how to go about this?

Answer 1

You need to do it explicitly. I normally do this with a wrapper function like this.

Edit: Now there’s a complete working example at the bottom.

struct mystruct *mystruct_init()
{
    struct mystruct *mystruct = calloc(1, sizeof(*mystruct));

    // loop through and allocate memory for each element in list
    for (int i = 0; i < 10; i++) {
        mystruct->list[i] = calloc(1, sizeof(*(mystruct->list[i])));
    }

    return mystruct;
}

That way, you just call this in your code:

struct mystruct *h1 = mystruct_init();

You’ll also want to write a corresponding mystruct_free() function.

Here’s a working example (for me):

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

struct y {
    int a;
};

struct mystruct {
    int x;
    struct y **list;
    int list_length;
};

struct mystruct *mystruct_init()
{
    struct mystruct *mystruct = calloc(1, sizeof(*mystruct));

    // loop through and allocate memory for each element in list
    mystruct->list_length = 10;
    mystruct->list = calloc(1, sizeof(struct y *) * mystruct->list_length);
    for (int i = 0; i < mystruct->list_length; i++) {
        mystruct->list[i] = calloc(1, sizeof(struct y));
    }

    return mystruct;
}

void mystruct_free(struct mystruct *mystruct)
{
    for (int i = 0; i < mystruct->list_length; i++) {
        free(mystruct->list[i]);
    }

    free(mystruct->list);
    free(mystruct);
}

int main(int argc, char *argv[])
{
    struct mystruct *h1 = mystruct_init();
    h1->x = 6;
    printf("%d\n", h1->x);
    mystruct_free(h1);
    return 0;
}