Append a node in a linkedlist - why segmentation error? - c

Question

I am implementing a linked-list in C with structure
struct node
{
int data;
struct node *next;
};
I have written the append function to add a node at the end of a linked-list, as below, and display function to display all the nodes. But display is giving segmentation fault due to some inconsistency in the append function I think. What can be wrong here? My book does have a similar function for append, using malloc. I want to know what is wrong in my function.
void append(struct node **q, int d)
{
struct node *temp;
temp = *q;
printf("\nBegin: Address at temp = %u", temp);
while (temp!= NULL){
temp = temp->next;
printf("\nTravel: Address at temp = %u", temp);
}
struct node p1;
p1.data = d;
p1.next = NULL;
*q=&p1;
printf("\nEnd: Address at *q = %u\n", *q);
printf("\n*q->data = %d next = %u", (*q)->data,(*q)->next );
}
void display(struct node *q)
{
printf("\n");
while (q != NULL){
printf(" -> %d",q->data);
q = q->next;
}
}
int main(int argc, char *argv[])
{
struct node *p;
p = NULL; /* empty linked list */
printf("\nNo. of elements in the Linked List = %d", count(p));
append(&p,1);
display(p);
append(&p,2);
display(p);
printf("\nNo. of elements in the Linked List = %d", count(p));
}
Output:
No. of elements in the Linked List = 0
Begin: Address at temp = 0
End: Address at *q = 6684096
*q->data = 1 next = 0
-> 1Segmentation fault
However, when I replace
struct node p1;
p1.data = d;
p1.next = NULL;
*q=&p1;
with
temp = *q;
*q = malloc(sizeof(struct node))
temp->data = d;
temp->next = NULL;
the error is gone.
Can someone explain the reason?

Answer 1

Here you use an address of a local variable in a way which makes it accessable after you left the function:
*q=&p1;
Then you leave the function.
Whenever that is accessed later, it will access memory which is NOT the local variable anymore.
You need to allocate the memory for the variable. Use malloc() for that.
E.g.:
struct node *p1;
p1 = malloc(sizeof(*p1));
/* skipping recommended check of success/NULL */
p1->data = d;
p1->next = NULL;
*q=p1;
There are more problems, like you let the list start with the new node, which is followed by NULL, which loses/leaks all of your previous list. But the immediate problem is caused by referencing the memory location of a bygone local variable after it exists.

Answer 2

regarding the loop:
while (temp!= NULL){
This results in temp containing NULL so this has stepped all the way through the linked list and off the end of the list.
Suggest:
while (temp->next != NULL){
as this will stop stepping through the linked list when it is pointing to the last 'node' in the linked list.
Then need to use temp as the pointer to the last 'node' in the linked list (where you want to append a new node)
regarding:
struct node p1;
p1.data = d;
p1.next = NULL;
*q=&p1;
this creates the new 'node' on the stack. However, anything on the stack 'disappears' when the function returns.
q is a pointer to the first 'node' in the linked list, not the last 'node' in the linked list. Suggest using temp as (after correcting 1) points to the last 'node' in the linked list
each 'node' needs to be created in the 'heap' memory, via malloc() or calloc() so it still exists after the function exits.
regarding:
temp = *q;
*q = malloc(sizeof(struct node))
temp->data = d;
temp->next = NULL;
the error is gone.
NO, the error is NOT gone. Rather, this always inserts the new 'node' as the second 'node' in the linked list. (and breaks the link to the following nodes of the linked list.)
the posted code fails to return all the allocated memory to the heap (via calls to free() ) The result is a memory leak for each and every call to malloc().