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?
while(root!=NULL){root is never changed inside this print-loop. BTW: this is a good reason to prefer for() loops over while() loops.My logic is correct but somehow code has a bug.That's not how it works. At a first glance, your printList function doesn't walk the list, but only prints root->data.insertreturns the head pointer, e.h:struct node *insert(struct node* root, int data) ...