CODEWITHSUNDEEP
cpointerslinked-list
Leet
Leet

Reputation: 83

Linked list as an argument to a function

The program does not print the list's values as it is intended to. It prints something that's gotta be a memory address imo. I've been trying to find the solution solo, but to no avail thus far. I would appreciate some help.

#include <stdio.h>

typedef struct node
{
    int val;
    struct node * next;
} node_t;

void print_list(node_t * head);

void main()
{
    node_t * head = NULL;
    head = malloc(sizeof(node_t));
    if (head == NULL)
        return 1;
    head->val = 1;
    head->next = malloc(sizeof(node_t));
    head->next->val = 2;
    head->next->next = malloc(sizeof(node_t));
    head->next->next->val = 3;
    head->next->next->next = malloc(sizeof(node_t));
    head->next->next->next->val = 18;
    head->next->next->next->next = NULL;

    print_list(&head);
    system("pause");
}

void print_list(node_t * head) {
    node_t * current = head;

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

The aforementioned problem was resolved thanks to your inputs. Thank you very much! However, a new issue occurred. Wanting to add a new element to the list, I added a few lines of code. Unfortunately, the wanted result is not printed and the program is terminated abruptly. Here's the new code:

    head->next->next->next->next = malloc(sizeof(node_t));
    head->next->next->next->next->val = 5556;
    head->next->next->next->next->next = NULL;
    node_t * current = head;
    while (current->next != NULL) 
    {
        current = current->next;
    }
    current->next = malloc(sizeof(node_t));
    current->next->val = 32;
    current->next->next = NULL;
    printf("%d\n", current->next->val);
    system("pause");
}

Upvotes: 0

Views: 2626

Answers (3)

C. Hubbard
C. Hubbard

Reputation: 11

Your print_list function is called using print_list(&head); , however head is already a pointer to a node.

This means that the print_list function is actually receiving a pointer to a pointer to a node, or node_t **. As such, when you try to print the values of the pointer you will see something that appears to be a memory address. To fix this, just pass in the head of the list using print_list(head);

Upvotes: 0

Rajeshkumar
Rajeshkumar

Reputation: 798

Change your print_list(&head) function to print_list(head).

print_list(&head);

to

print_list(head);

function void print_list(node_t * head) accepts single reference pointer. But your passing **head to the print_list function.

Upvotes: 0

GAURANG VYAS
GAURANG VYAS

Reputation: 689

Note that your function void print_list(node_t * head); expects a parameter of type node_t * but you are passing parameter of type node_t **.

Change print_list(&head); to print_list(head);

head is of type node_t * while &head is of type node_t **.

Upvotes: 5

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