CODEWITHSUNDEEP
c
Despicable.Me
Despicable.Me

Reputation: 193

2 ways of initializing a linked list, do they equal?

typedef struct node{
    int data;
    struct node* next;
}ListNode;

void init(ListNode **head){
    (*head) = (ListNode *)malloc(sizeof(ListNode));
    (*head)->next = 0;
}

ListNode* another_init(){
    ListNode *head = (ListNode *)malloc(sizeof(ListNode));
    return head;
}

I have some problems:

1.in function init,why should I put a second rank pointer ?

2.Is function init the same with another_init ?

Upvotes: 0

Views: 104

Answers (2)

Lidong Guo
Lidong Guo

Reputation: 2857

2.Is function init the same with another_init ?

In case you use malloc they are not equal If you use calloc ,they will be equal.

1.in function init,why should I put a second rank pointer ?

If you insist on malloc . You should set the next to 0 .

Or you will face bugs sometime. Thinks this:

Maybe you have a print_link function:

void print_link(struct node * head)
{
    node * now = head;
    while(now != NULL)
    {
        print_some_data_you_want;
        now = now->next;
    }
}

And if you just call this function after use the second init function. You never know what will happened.

Upvotes: 0

M.E.L.
M.E.L.

Reputation: 613

another_init is not "the same as" init. It doesn't set the next-pointer to 0. From malloc(3): malloc() allocates size bytes and returns a pointer to the allocated memory. The memory is not cleared.

Upvotes: 1

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