CODEWITHSUNDEEP
clinked-listmalloc
user11308707
user11308707

Reputation: 1

why can't I initialize a linked list like this?

LinkedList.c

#include "stdio.h"
#include "stdlib.h"


struct ListNode
{
    int val;
    struct ListNode *next;
};

int main()
{
    int a[] = {1, 2, 3};
    struct ListNode *head = malloc(sizeof(struct ListNode));
    head->next = NULL;
    struct ListNode *rear = head;
    for (int i = 0; i < 3; i++)
    {
        struct ListNode s;
        s.val = a[i];
        s.next = rear->next;
        rear->next = &s;
        rear = &s;
    }
    struct ListNode *tmp = head;
    head = head->next;
    free(tmp);


    //display
    while (head != NULL)
    {
        printf("%d ",(*head).val);
        head = head->next;
    }

    return 0;
}

If I change the sentece "struct ListNode s;" to "struct ListNode *s=malloc(sizeof(struct ListNode));" and use "->" but not ".", it will work.

Upvotes: 0

Views: 50

Answers (1)

Rohan Kumar
Rohan Kumar

Reputation: 5882

As pointed out in comments, you're facing this because ListNode you create in your list creation loop is a local variable limited to loop scope. It only exists in that scope(for that particular iteration). You should use malloc for allocating ListNode and use free to de-allocate memory for your nodes when you're done with using the list. I did minor changes to your list creation loop:

for (int i = 0; i < 3; i++) {
    struct ListNode *s = malloc(sizeof(struct ListNode));
    s->val = a[i];
    s->next = rear->next;
    rear->next = s;
    rear = s;
}

When I ran this, I was able to see expected results:

src : $ gcc initializelinkedlist.c 
src : $ ./a.out 
1 2 3

Upvotes: 1

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