How to print Linked List in C? - c

Question

#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
int data;
struct Node* next;
} Node, *LinkedList;
void CreateList(LinkedList N, int n)
{
N = (LinkedList)malloc(sizeof(Node));
N->next = NULL;
LinkedList new = N;
Node *p;
for (int i = 0; i < n; ++i) {
p = (Node *)malloc(sizeof(Node));
scanf("%d", &(p->data));
new->next = p;
new = p;
}
new->next = NULL;
}
int main()
{
LinkedList list;
CreateList(list, 20);
printf("%d", list->data);
return 0;
}
As you can see, I want to create a linkedlist and make it a function.
But when I "printf" linkedlist's data, it can't appear what i want.
Can you help me?

Answer 1

The direct problem, as M. Oehm notes, is that you pass the list object to the create function. The create function creates the list, but because the list object is not returned to main, main cannot see the list. To achieve what you want, do:
In main, declare the list as:
LinkedList *N; // a pointer
declare create as:
void CreateList(LinkedList **N, int n) // address of a pointer that receives the value
and dereference it in create:
*N = malloc(sizeof(Node)); // assign the value to the pointer in main
and now call it from main as:
CreateList(&N, 20); // pass the address of the pointer
I further note that you pass create an int, the number of elements in the list, but a list is typically made for an unknown number of elements. So you should read until end-of-file.
(All other required modifications in create I leave to you.)

Answer 2

Thank you all!! I solved this problem, and this is my code.
#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
int data;
struct Node* next;
} Node;
void create(Node* *head, int n)
{
*head = malloc(sizeof(Node));
(*head)->next = NULL;
Node* new = *head;
Node* p;
for (int i = 0; i < n; ++i) {
p = malloc(sizeof(Node));
scanf("%d", &(p->data));
new->next = p;
new = p;
}
}
int main()
{
Node* list;
create(&list,20);
printf("%d", ((list->next)->next)->data); //for test
return 0;
}