Pointers issue in linked list function's return I cannot figure

Question

When debugging it, it tells me L is nullptr. I can't figure out why it doesn't returns the list properly.

These are the structs (I must use list and node):

typedef struct node node;
typedef struct List list;
struct node {
    int data;
    node *next;          
};

struct List {
    node *head;
};

The function that creates the list:

void BuildList(list *L) {
    node *head = NULL, *temp = head;
    int num;
    printf("Input list's elements: \n");
    do {
        scanf("%d", &num);
        if (num != -1) {
            if (head == NULL) {
                head = BuildNode(num);
                temp = head;
            }
            else {
                temp->next = BuildNode(num);
                temp = temp->next;
            }
        }

    } while (num != -1);

    L = (list *) malloc(sizeof(list));
    L->head = head;
}

Auxiliary function for BuildList:

node* BuildNode(int num1) {
    node *node1 = (node *)malloc(sizeof(node));

    node1->data = num1;
    node1->next = NULL;

    return node1;
}

Printing function:

void PrintList(list *L) {
    node *head;
    head = L->head;
    printf("The list's elements are: ");

    while (head != NULL) {
        printf("%d ", head->data);
        head = head->next;
    }
    printf("\n");
}

The program fails on "head = L->head;" at PrintList, claiming it's a nullptr. It's origin is provably the dynamic allocation in BuildList, at the end. The calls from main are:

list *head = NULL;
BuildList(&head);
PrintList(head);

When replacing PrintList(head); with PrintList(&head); it prints an empty list, without failing.

which head? In BuildList or main? In BuildList head is reffered as L. — גל זיגלר
– גל זיגלר, Commented Dec 10, 2016 at 13:15

koper89 · Accepted Answer · 2016-12-10 13:21:53Z

2

You're passing a pointer to function:

BuildList(list *L)

Which means that when you're allocating it inside of function, you will not have this pointer outside of this function, because it's on stack. What you could do is either, allocate List outside of this function like:

list *head = malloc(sizeof(list)); /* It's a good habit to not cast malloc function */ 
BuildList(head); /* Remember to remove malloc from inside of build list */
PrintList(head);

Or pass the double pointer to function:

void BuildList(list **L) {
    node *head = NULL, *temp = head;
    .....
    *L = malloc(sizeof(list));
    (*L)->head = head;
}

list *head = NULL;
BuildList(&head);
PrintList(head);

answered Dec 10, 2016 at 13:21

koper89

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גל זיגלר Over a year ago

Thanks. It was helful.

ymdatta · Accepted Answer · 2016-12-14 19:21:06Z

0

Here you need to send address of the pointer to list l as arguement not the address of list,because we want all changes done in the function to affect list l,this can be achieved only if we have the address of the list in the function,so that any changes to list are changed permanently in the memory!

so u just have to use

void BuildList(list **lreference)

and just send the address of pointer to list while calling the BuildList function.

list *l;
l = malloc(sizeof(list));
BuildFirst(&l);

answered Dec 14, 2016 at 19:21

ymdatta

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