Correctly Implementing a Linked List in C - c

Question

I am trying to implement a linked list from scratch in C:
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node * next;
};
void insert(struct node** root, int data){
// Create a Node
struct node * temp = malloc(sizeof(struct node));
temp->data = data;
temp->next = NULL;
// Either root is NULL or not
if (*root == NULL){
*root = temp; // Directly modify the root
}
else {
struct node * t = *root;
while (t->next!=NULL){
t = t->next;
}
t->next = temp; // Append at the last
}
}
void printList(struct node * root){
while(root!=NULL){
printf("%d\t", root->data);
}
}
struct node * search(struct node* root, int key){
while (root!=NULL) {
if (root->data == key) return root;
}
return NULL;
}
int main(){
struct node * head = NULL;
insert(&head,0);
insert(&head,1);
insert(&head,2);
insert(&head,3);
insert(&head,4);
printList(head);
}
Now, when I run the program, my output is:
0 0 0 0 0 0 0 0 0 0
However, my list doesn't contain all zeroes or 10 elements.
My logic seems correct but somehow code has a bug.
On a side note, is there a way to avoid double pointers, can't I work with only pointers while inserting in a linked list?

Answer 1

There is a small bug in the printList() function.
In printList() function, root not updated, to iterate whole list you should do root = root->next
void printList(struct node * root){
while(root!=NULL){
printf("%d\t", root->data);
root = root->next; /* you miss this one */
}
}
Same mistake is repeated in search() function also,
struct node * search(struct node* root, int key){
while (root!=NULL) {
if (root->data == key)
return root;
else
root = root->next; /* if key not found root should be updated to next one */
}
return NULL;
}