recursively call in c to create link list - c

Question

I am new in c so any help will be appreciated. I need to print 10 numbers from a linked list (it doesnt matter which numbers for now) I believe my code will print 9,8,7...0. for example. The linked list will be part of a struct (struct data) that will contain other variables (not important for now)
//linked list
struct listOfNodes {
struct node *root;
};
//list of parameters to send to the function to print nodes
struct data {
struct listOfNodes *list;
int total;
};
I need to send the struct (struct data) as a parameter of a recursive function (addNode). In this recursive function, I need to add a new node to the linked list and call recursively 10 times to create more nodes for the link list, then I need to print the linked list. I have the following code so far
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
//node
struct node {
int value;
struct node *next;
};
//linked list
struct listOfNodes {
struct node *root;
};
//list of parameters to send to the function to print nodes
struct data {
struct listOfNodes *list;
int total;
};
void printNode(struct listOfNodes *list) {
struct node *n = list->root;
while(n!=NULL){
printf("%d\n",n->value);
n=n->next;
}
}
void addNode(void* d){ //parameter needs to be a void*
struct data *o = (struct data *)d ;
if(o->total<10) {
//create new node
struct node *n = malloc(sizeof(struct node));
n->value = o->total;
n->next = NULL;
o->total = o->total + 1;
if(o->list->root == NULL)
{
o->list->root = n;
}
else {
n->next = o->list->root->next;
o->list->root->next = n;
}
addNode(d);
}
}
int main() {
struct data *d= malloc(sizeof(struct data *));
d->total=0;
d->list=NULL;
addNode(d); //add recursively 10 times
if(d->list!=NULL) printNode(d->list);
return 0;
}
But I am getting Segmentation fault (core dumped). Can you please help me?

Answer 1

In your main program, you added list as NULL. But in your addNode, you only check if list->root is NULL. What's happening is when
if(o->list->root == NULL)
{
o->list->root = n;
}
is accessing list->root when list is NULL. You de reference a NULL pointer and segfault.
You probably need
struct listOfNodes *variable=malloc(sizeof(struct listOfNodes));
d->list=variable;

Answer 2

Changes are describe in comments
#include <stdio.h>
// #include <string.h> // <------- 4 not needed.
#include <stdlib.h>
struct node {//node
int value;
struct node *next;
};
struct listOfNodes {//linked list
struct node *root;
};
struct data {//list of parameters to send to the function to print nodes
struct listOfNodes *list;
int total;
};
void printNode(struct listOfNodes *list) {
struct node *n = list->root;
while(n!=NULL){
printf("%d\n",n->value);
n=n->next;
}
}
void addNode(struct data * d){ // <------- 3 no need to be void * so o replaced by d
if(d->total<10) {
//create new node
struct node *n = malloc(sizeof(struct node));
n->value = d->total;
n->next = NULL;
d->total = d->total + 1;
if(d->list->root == NULL)
{
d->list->root = n;
}
else {
n->next = d->list->root->next;
d->list->root->next = n;
}
addNode(d);
}
}
int main() {
struct data *d= malloc(sizeof(struct data)); // <---- 1 allocate data size not pointer size.
d->total=0;
d->list = malloc(sizeof(struct listOfNodes)); // <---- 2 alocate list !
d->list->root = NULL; // <---- 2 init root.
addNode(d); //add recursively 10 times
if(d->list!=NULL) printNode(d->list);
return 0;
}