C pointer assignment in array

Question

typedef struct {
    int head
} List;

List* array[10];
int size = 0;

List* create() {

    if(size < 10){

        List list1;
        list1.head = size;
        array[size] = &list1;
        printf("%d \n", array[size]->head); //value is stored perfectly here
        printf("%p \n", array[size]);       //same value printed everytime (e.g 0x7ffeee1139f0 )
        return array[size++];
    }
    return NULL;
}

int main() {
    List* plist1 = List_create();
    printf("%p \n", plist1);//same value 0x7ffeee1139f0

    List* plist2 = List_create();
    printf("%p \n", plist2);//same value 0x7ffeee1139f0
}

For the following code above, when I try to iterate over the array and print the value at each index, which should be the address of each list struct I created and hence should be unique.

But I get the same address stored at each index of the array. I can't figure out why.

Also when I iterate over the array and print array[index]->head, I get some garbage value(e.g. -477361872).

Does anyone know what am I doing wrong?

@Inian Sorry, I think it was a typo in my question. I corrected it. Thanks for pointing it out. — Aryan Arora
– Aryan Arora, Commented May 20, 2020 at 18:03

Alex · Accepted Answer · 2020-05-20 17:56:31Z

1

It looks like you are storing the address of a variable on the stack into the array. There is only one stack variable (list1) and so its addresss is the same every time. Or it might be different, but there is really ever only one valid list1 object.

You also should not use stack addresses after they have gone out of scope. So as soon as you leave the create function, the addresses you have stored in your array are invalid.

If you want to create an array of ten structs, you can do List array[10];. If you want to dynamically allocate each struct, you can use List *array[10]; and then assign into the array pointers you have allocated using new or malloc (or other valid pointers).

answered May 20, 2020 at 17:56

Alex

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2 Comments

Aryan Arora Over a year ago

Thanks for the response, I could but I am limited to not use malloc or new.

Alex Over a year ago

Are you able to use a stack allocated array of List objects, e.g. List array[10];? What are you trying to do?

August Karlstrom · Accepted Answer · 2020-05-20 18:56:56Z

0

One of the loopholes of the C programming language is that you can return a pointer to a local variable from a function. This is what is happening in the function create. The next time the function is called the value of the variable list1 is undefined. If you cannot use dynamic memory allocation, why not simply declare an array of lists instead of list pointers?

answered May 20, 2020 at 18:56

August Karlstrom

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Comments

Vlad from Moscow · Accepted Answer · 2020-05-20 18:57:34Z

The both calls of the function create stores the address of the same local variable list1 in the array array and return this address to the caller.

List* create() {
    List list1;
    //...
    array[size] = &list1;
    //..
    return array[size++];
    //...
}

So the both pointers in main plist1 and plist2 have the same invalid values because the local variable list1 is not alive outside the function create.

You should allocate memory for objects of the type List dynamically.

For example

List* create() {

    if(size < 10){

        List *list1 = malloc( sizeof( List ) );

        if ( list1 != NULL )
        {
            list1->head = size;
            array[size] = list1;
            printf("%d \n", array[size]->head);
            printf("%p \n", ( void * )array[size]);
            return array[size++];
        }
    }

    return NULL;
}

In this case you should not forget to free the allocated memory when it will not anymore needed.

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