This a follow up from this question: Why does a function which returns void not hold value?. A user indeed suggested me to draw the pointers in the function struct node * insert_front and so the main questions is about that function and this line of code head = new
This is the minimum code needed:
#include <stdio.h>
#include <stdlib.h>
struct node {
int value;
struct node* next;
};
struct node* init(int value) {
struct node* head = malloc(sizeof(struct node));
if (!head) return NULL;
head->value = value;
head->next = NULL;
return head;
}
void insert_back(struct node* head, int value) {
if (!head) return;
struct node* temp = head;
struct node* new = malloc(sizeof(struct node));
while (temp->next != NULL) {
temp = temp->next;
}
temp->next = new;
new->value = value;
new->next = NULL;
return;
}
struct node* insert_front(struct node* head, int value) {
struct node* new = (struct node*)malloc(sizeof(struct node));
new->value = value;
new->next = head;
printf("head->value: %d\n", head->value);
printf("head->value ptr: %p\n", &head->value);
printf("new->value: %d\n", new->value);
printf("new->value ptr: %p\n", &new->value);
head = new;
printf("head->value: %d\n", head->value);
printf("head->value ptr: %p\n", &head->value);
printf("new->value: %d\n", new->value);
printf("new->value ptr: %p\n", &new->value);
return new;
}
int main() {
struct node* head = init(5);
insert_back(head, 7);
insert_back(head, 9);
insert_back(head, 12);
head = insert_front(head, 999)
return 1;
}
The output produced from struct node* insert front(...) is
// after, new->next = head
head->value: 5
head->value ptr: 0x(...)9e0
new->value: 999
new->value ptr: 0x(...)a50
// after, head = new
head->value: 999
head->value ptr: 0x(...)a50
new->value: 999
new->value ptr: 0x(...)a50
So we see that after head = new we have the same addresses for head and new. Given this I have two questions.
Firstly, have I understood correctly that, after head = new what happens is that we have moved the pointer which was pointing to the start of that region of memory used by the linked list (that is from head) to a new place in memory (that is where new is in memory)? Otherwise (or more technically, well-explained) what happened?
Secondly, when I print:
printf("&head->value: %p\n", &head->value);
printf("&head: %p\n", &head);
printf("head: %p\n", head);
I obtain
&head->value: 0x(...)9e0
&head: 0x(...)8c8
head: 0x(...)9e0
So, with head I am accessing the address of the region of memory I am pointing to and with &head the address of the pointer itself. But then why with &head->value I am accessing the same as with head?
