Linked list implementation in C(printing only last two nodes) - c

Question

#include <stdlib.h>
#include <stdio.h>
struct node {
int data;
struct node *next;
};
void addLast(struct node **head, int value);
void printAll(struct node *head);
struct node *head1 = NULL;
int main() {
addLast(&head1, 10);
addLast(&head1, 20);
addLast(&head1, 30);
addLast(&head1, 40);
printAll(head1);
return 0;
}
void addLast(struct node **head, int value) {
struct node *newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = value;
if (*head == NULL) {
*head = newNode;
(*head)->next = NULL;
} else {
struct node **temp = head;
while ((*temp)->next != NULL) {
*temp = (*temp)->next;
}
(*temp)->next = newNode;
newNode->next = NULL;
}
}
void printAll(struct node *head) {
struct node *temp = head;
while (temp != NULL) {
printf("%d->", temp->data);
temp = temp->next;
}
printf("\n");
}
addLast() will append the new node at the end of the list, with printAll(), I am printing entire list.
Every time when I am printing the list, I can only see the last two nodes.
Can anyone please help, why loop is not iterating over entire list ?

Answer 1

The function addLast is too complicated and as result is wrong due to this statement
*temp = (*temp)->next;
in the while loop. It always changes the head node.
Define the function the following way
int addLast( struct node **head, int value )
{
struct node *newNode = malloc( sizeof( struct node ) );
int success = newNode != NULL;
if ( success )
{
newNode->data = value;
newNode->next = NULL:
while( *head ) head = &( *head )->next;
*head = newNode;
}
return success;
}
Take into account that there is no need to declare the variable head1 as global. It is better to declare it inside the function main.
Also all the allocated memory should be freed before exiting the program.