Why the appending in C linked list failed - c

Question

I am new to C, now I am making a linked list for face detection.
Below is the struct and the method for appending face at the end of linked list.
//Structure for storing a face with x, y and window size
typedef struct Face {
int window;
int x;
int y;
struct Face* next;
} Face;
//Append face(window, x, y) to the end of linked list starting from head
void push(Face* head, int window, int x, int y) {
Face* temp = (Face *)malloc(sizeof(Face));
temp->window = window;
temp->x = x;
temp->y = y;
temp->next = NULL;
Face* cur = head;
if (head == NULL) {
printf("Called\n");
head = temp;
} else {
while (cur->next != NULL) {
cur = cur->next;
}
cur->next = temp;
}
}
In another file, the executable, I called push(head, 1, 2, 3)[head here is initialized to NULL].
Only "Called' is printed on the screen. And the head is still NULL in the executable when I examine the linked list.
I have no idea why the head is not updated, but when I have it in the same file, it seems to work fine.

Answer 1

It's a guessing game since you don't show the relevant code.
Luckily, it's quite easy to guess well in this case...
The parameter you pass into the function is of type Face * and you set it to a new value (the new struct you allocated). Unfortunately, you're not returning this value, nor are you making sure the input parameter is capable of "transferring" data back to the calling context. What you should do is:
void push(Face** head, int window, int x, int y) {
// all you code here...
*head = temp
// rest of code...
}
And when you call the function:
push(&head, 1, 2, 3);

Answer 2

if (head == NULL) {
printf("Called\n");
head = temp;
}
The assignment head = temp only modifies the local copy of the head pointer. So it is not propagated to the code that called push(). If head was NULL in the code that calls push(), then it will remain so.
You could for example return the list head, as in:
Face *push(Face* head, int window, int x, int y) {
Face* temp = (Face *)malloc(sizeof(Face));
temp->window = window;
temp->x = x;
temp->y = y;
temp->next = NULL;
Face* cur = head;
if (head == NULL) {
printf("Called\n");
head = temp;
} else {
while (cur->next != NULL) {
cur = cur->next;
}
cur->next = temp;
}
return head;
}
And then use it like this:
/* ... */
head = push(head, window, x, y);
/* ... */
Another alternative is to pass a pointer to pointer to head (Face **), and replace that assignment with *head = temp;, but if you're a beginner I'd stick to the previous approach (and in my opinion, using double indirection is not necessary in most of the cases, but this can be subjective).
Finally, you might want to handle possible malloc(3) errors: the allocation can fail, you should check and handle that case.

Answer 3

I believe it is because you are passing in the value of the pointer head, which creates a copy of the pointer. By setting head to another address, you're not modifying head outside of scope, but rather the head within the method scope.
You'd need to pass in a pointer to the pointer to change it.

Answer 4

Function parameters are local variables of function. They are copies of the arguments. So the function parameter head is a copy of the argument head. Any changes of the parameter (of the copy of the argument) does not influence on the argument. You have to pass the head by reference.
Define the function the following way
//Append face(window, x, y) to the end of linked list starting from head
void push( Face **head, int window, int x, int y )
{
Face *temp = malloc( sizeof( Face ) );
if ( temp != NULL )
{
temp->window = window;
temp->x = x;
temp->y = y;
temp->next = NULL;
while ( *head ) head = &( *head )->next;
*head = temp;
}
}
And call the function like
push( &head, 1, 2, 3 );