C linked list to Display contents - c

Question

Is there any part i have missed in the code?
I am creating a non-empty linked
list and to display the contents of a linked list. Where am I getting it wrong?
#include <stdbool.h>
#include <stdlib.h>
struct node_int
{
void *data;
node next;
};
typedef struct node_int *node;
typedef struct list_int { node first; } *list;
void init_list(list *lp, void *o)
{
*lp = (list) malloc(sizeof(struct list_int));
(*lp)->first = NULL;
(*lp)->first->data = o;
(*lp)->first->next = NULL;
}
void print(list ell, void (*print_data)(void *d))
{
list c;
c = ell;
while (c!NULL)
{
print_data(c->data);
c = ell;
}
}

Answer 1

There are a few problems with your code.
First like to say that I find it bad style to typedef a pointer. If you do that, you should at least use a name which clearly tells that the type is a pointer. Names like list and node make others think of something which is not pointers.
Below is some code showing how it could look without typedef'ed pointers.
#include <stdio.h>
#include <stdlib.h>
struct node_int
{
void *data;
struct node_int* next;
};
typedef struct node_int node;
typedef struct list_int { node* first; } list;
void init_list(list** lp, void *o)
{
// Allocate the list
*lp = malloc(sizeof(list));
if (*lp == NULL) return;
// Allocate the first node
(*lp)->first = malloc(sizeof(node));
if ((*lp)->first == NULL)
{
free(*lp);
*lp = NULL;
return;
}
// Initialize first element
(*lp)->first->data = o;
(*lp)->first->next = NULL;
}
void print(list* ell, void (*print_data)(void *d))
{
if (ell == NULL) return;
node* p = ell->first;
while (p != NULL)
{
print_data(p->data);
p = p->next;
}
}
void myPrint(void* d)
{
int* p = (int*)d;
printf("%d\n", *p);
}
void free_list(list* ell)
{
// Add code here ...
}
int main(void)
{
int a = 1;
list* myList;
init_list(&myList, &a);
if (myList == NULL) return 0;
// Use the list.....
print(myList, myPrint);
free_list(myList);
return 0;
}