CODEWITHSUNDEEP
cpointers
saurabh agarwal
saurabh agarwal

Reputation: 2184

Error in Link List, pointer

what is wrong with this code? This is simply creating a link list and printing it.

#include<stdio.h>
#include<stdlib.h>

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

void printLinklist(struct node *head)
{
    struct node *temp = head;
    while(temp != NULL)
    {
        printf("%d\t",temp->data);
        temp = temp->next;
    }
    printf("\n");
}

void pushData(struct node *head, int val)
{
    struct node *temp = malloc(sizeof (struct node));
    temp->next = head;
    temp->data = val;
    head = temp;
}

int main()
{
    struct node *head = NULL;
    pushData(head, 1);
    pushData(head, 2);
    pushData(head, 3);
    pushData(head, 4);
    pushData(head, 5);

    printLinklist(head);
}

I know correct way of doing it. For that we will have to pass &head in pushData() function and define argument of this function as double pointer (replace head to *head in pushData() function and passing &head as an argument)

I can't understand the problem in this code.

Upvotes: 0

Views: 34

Answers (3)

Sel&#231;uk Aydınlı
Sel&#231;uk Aydınlı

Reputation: 1

The True code is:

#include <stdio.h>
#include <stdlib.h>
#pragma warning(disable:4996)   
struct node {
    struct node* next;
    int data;
};

void printLinklist(struct node* head)
{
    struct node* temp = head;
    while (temp->next != NULL)
    {
        printf("%d\t", temp->data);
        temp = temp->next;
    }
    printf("\n");
}

struct node* pushData(struct node* head, int val)
{
    struct node* temp = malloc(sizeof(struct node));
    if (temp == NULL)
        return NULL;
    temp->next = head;
    temp->data = val;

    return temp;
}

int main()
{
    struct node* head = malloc(sizeof(struct node));

    head = pushData(head, 1);
    head = pushData(head, 2);
    head = pushData(head, 3);
    head = pushData(head, 4);
    head = pushData(head, 5);
    struct node* head2 = head;
    for (int i = 0; i < 5; i++)
    {
        head2 = head2->next;
    }
    head2->next = NULL;
    printLinklist(head);
}

Upvotes: 0

Arjun Sreedharan
Arjun Sreedharan

Reputation: 11453

Arguments are always passed by value. When calling pushData(), the value of head, that is the address stored in the pointer variable is copied into the local variable head of the function. Changing this won't change the head variable in the main() function.

If you need that to happen, you have to pass &head and then assign value by dereferencing it in the pushData() function.

Upvotes: 3

Iharob Al Asimi
Iharob Al Asimi

Reputation: 53016

The problem is that you assign NULL to head in main(), and that never chagnes, it could work like this

struct node *pushData(struct node *head, int val)
{
    struct node *temp = malloc(sizeof (struct node));
    if (temp == NULL)
        return NULL;
    temp->next = head;
    temp->data = val;

    return temp;
}

and in main

int main()
{
    struct node *head;

    head = pushData(head, 1);
    head = pushData(head, 2);
    head = pushData(head, 3);
    head = pushData(head, 4);
    head = pushData(head, 5);

    printLinklist(head);
}

passing the pointer only modifies the data the pointer points to, but not the pointer itself.

So in your pushData() function, you do this

head = temp;

which changes the local pointer, and makes it pointer to temp, but once the function returns, the change is lost.

When you pass the address of head like &head, and then you can alter the pointer because your head pointer will be pointing to the pointer itself instead of the data the pointer points to.

Upvotes: 1

Related Questions