Passing struct to function and adding elements - c

Question

I'm new to C. I'm trying to pass a struct list to a function and within that function fill the list. Code is as follows:
#include <stdio.h>
#include <stdlib.h>
struct Abc {
int test;
struct Abc *next;
};
void demo_fill(struct Abc *data);
int main(int argc, char **argv) {
struct Abc *db = NULL;
demo_fill(db);
printf("%d\n",db->test);
return 0;
}
void demo_fill(struct Abc *data) {
int i;
for( i = 0; i < 5; i++ ) {
struct Abc *new;
new = malloc(sizeof(struct Abc));
new->test = i;
new->next = data;
data = new;
}
}
When running this a 'Segmentation fault (core dumped)' error occurs because the struct is still NULL when I try to print the first element. What am I doing wrong?

Answer 1

You're passing the pointer by value. You need to pass a pointer to a pointer if you want the change the value of the caller's pointer:
int main(int argc, char **argv) {
struct Abc *db = NULL;
demo_fill(&db);
printf("%d\n",db->test);
return 0;
}
void demo_fill(struct Abc **data) {
int i;
for( i = 0; i < 5; i++ ) {
struct Abc *new;
new = malloc(sizeof(struct Abc));
new->test = i;
new->next = *data;
*data = new;
}
}

Answer 2

Assigning data to new will have no effect. data is a local copy of the pointer. Pass a double pointer to fix that. Something like this:
void demo_fill(struct Abc** data) {
int i;
for( i = 0; i < 5; i++ ) {
struct Abc *new;
new = malloc(sizeof(struct Abc));
new->test = i;
new->next = *data;
*data = new;
}
}
And of course you will have to pas the pointer to db in main:
demo_fill(&db)

Answer 3

You can pass a pointer to the pointer as the other two answers say, so i just point out an alternative you might also consider: Instead of using a single struct you could use one struct for the list and another for entry-links:
struct link {
int test;
struct link* next;
};
struct list {
struct link* first;
};
void demo_fill(struct list* data);
You can then modify the first entry of the list without thinking about ** syntax.