动态创建和初始化指向C语言中的struct的指针数组

Question

Here is an example code of what I'm trying to achieve

typedef struct
{
  int x_pos, y_pos;
} A;

typedef struct
{
  A **arrayAp;
} B;

int main(void) {
    B B;
    int n = 10;
    A *array[n];
    B.arrayAp = array;
    for (int i = 0; i < 10; i++)
    {
    B.arrayAp[i] = malloc(sizeof(A));
    B.arrayAp[i]->x_pos = i;
    B.arrayAp[i]->y_pos = i + 1;
    }
    printf("%d",B.arrayAp[0]->x_pos);
}

It works as I want it to work. I can access elements of "array" using "arrayAp" pointer. But when I try to move some of this code to function for example:

A * makeAnArray()
{
    int n = 10;
    A *array[n];
    return &array;
}

and then assign value that it returns to "arrayAp"

B.arrayAp = makeAnArray();

Now I can't access first element of the array. Program just crash.

printf("%d",B.arrayAp[0]->x_pos);

But starting from second element everything works the same.

But the bigger problem I get when I try to move to function this code:

void initializeAnArray(B *B)
{
    for (int i = 0; i < 10; i++)
    {
    B->arrayAp[i] = malloc(sizeof(A));
    B->arrayAp[i]->x_pos = i;
    B->arrayAp[i]->y_pos = i + 1;
    }
}

Seems like it has no effect. When i'm trying to access array member through "arrayAp" pointer i'm getting some "random" values from memory.

typedef struct
{
  int x_pos, y_pos;
} A;


typedef struct
{
  A **arrayAp;
} B;

A * makeAnArray()
{
    int n = 10;
    A *array[n];
    return &array;
}

void initializeAnArray(B *B)
{
    for (int i = 0; i < 10; i++)
    {
    B->arrayAp[i] = malloc(sizeof(A));
    B->arrayAp[i]->x_pos = i;
    B->arrayAp[i]->y_pos = i + 1;
    }
}

int main(void) {
    B B;
    B.arrayAp = makeAnArray();
    initializeAnArray(&B);
    printf("%d",B.arrayAp[1]->x_pos);
}

Sorry for my poor English.

Answer 1

A * makeAnArray()
{
    int n = 10;
    A *array[n];
    return &array;
}

You are returning a local pointer which dies after the function ends and causes UB.

Moreover return type of your function should be A** and you should allocate memory with *alloc and return to ensure every instance is different and is alive after the function call.

Possible fix

A * makeAnArray()
{
    int n = 10;
    A *array = malloc(n * sizeof(A));
    return array;
}

Answer 2

Your function should be:

A * makeAnArray()
{
    int n = 10;
    //A *array[n]; --> This is local to this function so won't work
    A *array = malloc(sizeof(A)*n); //--> Dynamic allocation
    return array;
}

Answer 3

The part

A * makeAnArray()
{
    int n = 10;
    A *array[n];
    return &array;
}

is not going to work. The array array might be allocated on the stack, it cannot be accessed via the return value after termination of makeAnArray ; the result is undefined behaviour.