Segmentation Fault on Linked List Using C - c

Question

I am new to C and try to learn how to implement C on linked list. I am really confused why I can't access myList in the main function? because when I try to myList->data, it's segmentation fault. I think there's some error in my addtohead function?
Below is my code:
#include <stdio.h>
#include <stdlib.h>
typedef struct NODE{
int data;
struct NODE *next;
}node;
node * myList;
node * addToHead(node*, int);
void printList();
int main(){
myList = NULL;
int input;
while (scanf("%i",&input) == 1){
addToHead(myList, input);
printf("%d \n", myList->data);
}
printf("My List:\n");
printList(myList);
return 0;
}
node* addToHead(node* head, int newData){
node *temp = (node *)malloc(sizeof(node));
temp -> data = newData;
temp -> next = NULL;
if(head != NULL){
temp -> next = head;
}
head = temp;
return head;
}
void printList(node* head){
node *temp = head;
while(temp != NULL){
printf("%d ", temp->data);
temp = temp -> next;
}
printf("\n");
}

Answer 1

Your addToHead function is supposed to return the mallocated memory back to caller.
So you should assign the return value to mylist at first:
int main(){
node *myList = NULL;
int input;
while (scanf("%i",&input) == 1){
myList = addToHead(myList, input);
printf("%d \n", myList->data);
}
printf("My List:\n");
printList(myList);
return 0;
}
Into your addToHead function you wrote
head = temp;
But head has local scope and the assigned value is not reflected to the pointer myList.
To do so you have to use pointer to pointer.
int main(){
node *myList = NULL;
int input;
while (scanf("%i",&input) == 1)
{
if (addToHead(&myList, input) == true)
{
printf("%d \n", myList->data);
}
else
{
fprintf(stderr, "Error addToHead\n");
}
}
return 0;
}
bool addToHead(node** head, int newData){
node *temp = malloc(sizeof(node));
if (temp != NULL)
{
temp -> data = newData;
temp -> next = NULL;
if(head != NULL)
{
temp -> next = *head;
}
*head = temp;
return true;
}
return false;
}
Finally always remember to check the malloc return value: it can fail.

Answer 2

You return the new head node from addToHead, but you don't do anything with it. You need to assign this value to myList to update it:
myList = addToHead(myList, input);
Also, you misspelled a variable on the following line:
printf("%d \n", myListd->data);
It should be:
printf("%d \n", myList->data);

Answer 3

In this function definition
node* addToHead(node* head, int newData){
node *temp = (node *)malloc(sizeof(node));
temp -> data = newData;
temp -> next = NULL;
if(head != NULL){
temp -> next = head;
}
head = temp;
return head;
}
The parameter node* head is a local variable of the function. Any changes of the parameter will not influence on the original argument. The function parameters will be destroyed after exiting the function.
You can consider the function definition and its call the following way
addToHead(myList, input);
//...
node* addToHead(/*node* head, int newData*/){
node *head = myList;
int newData = input;
node *temp = (node *)malloc(sizeof(node));
temp -> data = newData;
temp -> next = NULL;
if(head != NULL){
temp -> next = head;
}
head = temp;
return head;
}
So the original variable myList will not be changed after the function call. You have to assign explicitly the returned value to the variable
myList = addToHead(myList, input);
Also the function has a drawback. It does not report an error in case when the new node was not allocated.
A better approach to write the function looks the following way
int /* _Bool */ addToHead( node **head, int newData )
{
node *temp = ( node * )malloc( sizeof( node ) );
int /* _Bool */ success = temp != NULL;
if ( success )
{
temp -> data = newData;
temp -> next = *head;
*head = temp;
}
return success;
}
In this case the function can be called in a loop the following way
while ( scanf( "%i", &input ) == 1 && addToHead( &myList, input ) )
{
//...
}