why my my function is using call by value method? - pointers

Question

I don't know why in last line it is printing data of first element instead of last element. I want explanation.
// A simple C program for traversal of a linked list
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// This function prints contents of linked list starting from
// the given node
void printList(struct Node* n)
{
while (n != NULL) {
printf(" %d ", n->data);
n = n->next;
}
}
int main()
{
struct Node* head = NULL;
struct Node* second = NULL;
struct Node* third = NULL;
// allocate 3 nodes in the heap
head = (struct Node*)malloc(sizeof(struct Node));
second = (struct Node*)malloc(sizeof(struct Node));
third = (struct Node*)malloc(sizeof(struct Node));
head->data = 1; // assign data in first node
head->next = second; // Link first node with second
second->data = 2; // assign data to second node
second->next = third;
third->data = 3; // assign data to third node
third->next = NULL;
printList(head);
printf("%d",head->data);
return 0;
}
As the function is accepting pointers so it should be call by reference.
And in last loop of function when n pointer is equal to NULL.
But in last line of this code is printing data of first list of my linked list.

Answer 1

Actually what you are doing is not being done in the actual linked list, it not pass by reference
void printList(struct Node* n)
{
/* some code here */
}
void main()
{
/* all your code here */
printList(head);
}
so if you want to change the head in the actual linked list you will have to pass the address of the pointer head to the function
something like this
int append_list(node **head, int data)
{
while((*head)->next!=NULL)
{
(*head) = (*head)->next;
}
}
int main()
{
struct node *head = NULL;
/* add nodes */
print_list(&head);
}
so here is the modification in your code:
#include <stdio.h>
#include <stdlib.h>
struct Node {
int data;
struct Node* next;
};
// This function prints contents of linked list starting from
// the given node
void printList(struct Node** n)
{
while ((*n)->next != NULL) {
printf(" %d ", (*n)->data);
(*n) = (*n)->next;
}
}
int main()
{
struct Node* head = NULL;
struct Node* second = NULL;
struct Node* third = NULL;
// allocate 3 nodes in the heap
head = (struct Node*)malloc(sizeof(struct Node));
second = (struct Node*)malloc(sizeof(struct Node));
third = (struct Node*)malloc(sizeof(struct Node));
head->data = 1; // assign data in first node
head->next = second; // Link first node with second
second->data = 2; // assign data to second node
second->next = third;
third->data = 3; // assign data to third node
third->next = NULL;
printList(&head);
printf("%d",head->data);
return 0;
}
here the output will be
1 2 3
since you have used (*head) for the traversal you no longer have the access to your list and hence will get segmentation fault if you try to access
(*head)->next
But I would not suggest to do this since now you will not be able to deallocate the memory

Answer 2

There is no pass-by-reference in C, everything is pass-by-value. People use pointers to emulate pass-by-reference, and this works because you can use the passed-in pointer to get at the same underlying data item.
In other words, even though the passed-in pointer is a pass-by-value copy within the function, the fact that it has the same value as the original means that both point to the same thing.
However, if the thing you're trying to change is a pointer already, you need a pointer to a pointer to do this emulation.
I could give you the code to do this but, believe me, it's not want you want. It would mean that the list printing code would be destructive to the list itself, since the head would now point to NULL.
Here is some code instead which shows how to do something similar, one that uses this double-pointer method to change the pointer outside of the function:
#include <stdio.h>
#include <stdlib.h>
void allocateSomeMem(void **pPtr, size_t sz) {
*pPtr = malloc(sz);
}
int main(void) {
void *x = NULL;
printf("%p\n", x);
allocateSomeMem(&x, 42);
printf("%p\n", x);
}
You can see by the output that the pointer is being changed:
(nil)
0x55f9ce5f96b0
Now, obviously, you wouldn't do this for the simple example shown, it would be far easier just to return the new pointer and have it assigned to x. But this is just illustrative of the method to use.