How to implement singly linked list in C - c

Question

I am trying to populate my linked list in C, but id does not word how
I want to. I want to keep a pointer "p" and keep adding to the list with the same pointer. However when I try to print, it prints only the data of the head!
#include<stdio.h>
#include <stdlib.h>
typedef struct{
int data;
struct node *next;
}node;
int main(){
node *head = NULL;
head = malloc(sizeof(node));
if(head==NULL){
printf("ta foirer quelque chose frero!");
return 1;
}
(*head).data=3;
(*head).next=NULL;
node *p = NULL;
p = (node*) head->next;
p = malloc(sizeof(node));
p->data = 5;
p->next = NULL;
p= (node *)p->next;
int i=0;
while(i<5){
p = malloc(sizeof(node));
i++;
p->data = i;
p->next=NULL;
p= (node *)p->next;
}
p = head;
while(p){
printf("\n%d",p->data);
p =(node*) p->next;
}
return 0;
}
I am getting as output
3
and I am expecting to get
3
5
0
1
2
3
4

Answer 1

#include<stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node *next;
};
typedef struct Node node;
void insert(node* h, int v) {
node* tmp = h;
while(tmp->next)
tmp = tmp->next;
node* newnode = malloc(sizeof(node));
newnode->data = v;
newnode->next = NULL;
tmp->next = newnode;
}
int main(){
node *head = NULL;
head = malloc(sizeof(node));
if(head==NULL){
printf("ta foirer quelque chose frero!");
return 1;
}
head->data=3;
head->next = NULL;
node *p = NULL;
insert(head, 5);
int i=0;
while(i<5){
insert(head, i++);
}
p = head;
while(p){
printf("%d\n",p->data);
p = p->next;
}
return 0;
}
If you notice, I changed the layout of your code a bit, and made it cleaner. What you needed was to traverse to find the node that appears before the place to add the new node, which in this case is the end. Usually this is a separate pointer in a different struct that contains the head and this is called the tail of a linked list. You were simply not keeping track of where to really add the node. This insert function above does that.