How to create, modify, and return array of pointers in a function in C?

Question

Here's a very basic example of what I'm trying to do (note that this segmentation faults)

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

typedef struct foo {
    int *bar;
} Foo;

Foo **fooPointers() {
    Foo **test = (Foo**) malloc(sizeof(struct foo) * 3);
    for(int i = 0; i < 3; i++) {
        Foo *curr = *(test + i);
        int *num = &i;
        curr->bar = num;
    }
    return test;
}

int main() {
    fooPointers();
    return 0;
}

the goal is to create an array of pointers of Foo, give each element meaningful values, and then return the pointer array.

Is anyone able to point me in the right direction as to why this doesn't work and how I can accomplish this task?

Answer 1

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

typedef struct foo
{
    int *bar;
} Foo;

Foo **fooPointers()
{
    Foo **test = malloc(sizeof(Foo*) * 3);  // should be `sizeof(Foo*)`
    
    static int k[] = {0,1,2};  // new array

    for(int j=0;j<3;j++)
    {
        test[j] = malloc(3*sizeof(Foo));  // No need to cast output of malloc
    }

    for (int i = 0; i < 3; i++)
    {
        Foo *curr = *(test + i);
        //int *num = &i;
        curr->bar = &k[i]; // storing different addresses.
    }
    return test;
}

int main()
{
    Foo **kk;

    kk = fooPointers();

    for(int i=0;i<3;i++)
    {
        printf("%d\n", *(kk[i]->bar));  //printng the values.
    }
    return 0;
}

The output is :

0
1
2