Problem with pointers in C - c

Question

I have a problem with my pointers and structures in C (I know, I knooww, pretty basic!). I was practicing my procedural paradigm. It's the first time I use a debugger, because I haven't really needed it earlier in my life : < so I if you please help me I'll be thankful.
I defined the following structure to make a list:
typedef struct node {
int info;
struct node *next;
struct node *prev;
} node_t;
And then this function to fill it up:
void addNodo(node_t * list, int x){
node_t * pointer;
node_t * temp;
temp = (node_t *)malloc(sizeof(node_t));
temp->info = x;
temp->next = NULL;
temp->prev = NULL;
pointer = list;
if(pointer == NULL){ //it's empty so start it
list = temp;
return;
}
if (pointer->info <= x) { //I like my lists tidy so...
while((pointer->next != NULL) && (pointer->info <= x)){
pointer = pointer->next;
}
if(pointer->next == NULL){
pointer->next = temp;
temp->prev = pointer;
return;
}
pointer->next->prev = temp;
temp->next = pointer->next;
temp->prev = pointer;
pointer->next = temp;
return;
}
}
And then, doing this:
int main(int argc, char** argv) {
node_t * list = NULL;
addNodo(list, 1);
printf("x: %d", list->info);
return (EXIT_SUCCESS);
}
It's throwing me a Segmentation Error! When I debug it everything is fun and games until it passes the ++++ line, list address goes back to 0x0 and can't get it to work. I know there's an error somewhere, but to my knowledge of pointers, it's perfectly fine. Please, detect my error and teach me some pointers.

Answer 1

When you call addNode() you're passing in the pointer by value. So when you change it in the body of the function the change is lost and doesn't propagate outside the function. You need to declare it as:
void addNode(node_t **pointer, int x)
and then use *pointer in the function.
And when you call ity in main, pass in &list

Answer 2

The problem is that you cannot modify list inside the addNodo function. In C parameters are sent by value, so the changes you are doing inside "addNodo" is local to there.
You need to change addNodo function so, it actually receives the direction where is list.
void addNode(node_t **list, int x){
...
if(*pointer==NULL){
*list = temp;
}
}
Then in your main you should use:
addNode(&list, 1);

Answer 3

Well, you are making the mistake of passing the address of the list by value. So all the arguments of the function are made copies of and then your addNodo() works on the copied variables. Thus the original list does not get modified.
What you should be doing while calling is this:
addNodo(&list, 1);
In the function make these changes:
void addNodo(node_t ** list, int x)
/* This will enable you to get a copy of the address of the list variable.
Please note that this is also pass by value, C does not support pass by
reference */
Then make this change:
pointer = *list;
/* this will make the pointer point to the beginning of list as now
list is a pointer to pointer type */
Hope it helps you.
BTW, please go through a standard C book (I recommend K&R) to get familiar with passing arguments in C and what happens internally.

Answer 4

You're making a classic mistake:
void addNodo(node_t * list, int x)
...
list = temp;
return;
list isn't changed in the caller (main())
You can change the values in the memory list points at, but you can't change the value of list and have it be seen by the caller.
In order to do that, you'd need to pass a pointer to a pointer into the function:
void addNodo(node_t **list int x)
This allows you to change what list points at by doing:
*list = temp;