Linked List in C

Question

Good morning/afternoon/evening everyone !

https://code.sololearn.com/cdyGRqAQvZFs/?ref=app

I passed a reference to the pointer linked_list from main() function to show() function, then I changed its value (address).. I don't understand why, after the execution of show(), linked_list is not pointing to NULL (0) and still pointing to the head of the linked list.

This is the output:

Number of Elements: 3
Element 1: 5
Element 2: 8
Element 3: 2

BP1: linked_list is at 43f610
Linked list contains: 5 8 2
BP2: linked_list is at 0
BP3: linked_list is at 43f610

Can anyone please explain why ?
And thanks in advance :)

Answer 1

nihil Yes you did :)))) Thank you so much ! I also googled the use of a pointer to a pointer since this is new to me to better understand it and I came up with the following example:

#include<stdio.h>

int main() {
int n = 5, * p = &n, ** pp = &p;

printf("n = %d", n);
printf("\n&n = %d", &n);

printf("\n\np = %d", p);
printf("\n&p = %d", &p);
printf("\n*p = %d", *p);

printf("\n\npp = %d", pp);
printf("\n&pp = %d", &pp);
printf("\n*pp = %d", *pp);
printf("\n**pp = %d", **pp);

return 0;
}

The output is:

n = 5
&n = 2293324

p = 2293324
&p = 2293312
*p = 5

pp = 2293312
&pp = 2293304
*pp = 2293324
**pp = 5

Again.. Thank you so much, sir or madam, for your explanation. :)

Answer 2

nihil Thanks ! You too. Happy coding and have a nice day :)