After exiting from function, nodes aren't saved - c++

Question

My program should create a linked list and show it. My problem is when the addelemnt_end function ends, it doesn't update head and last.
I tried with debug and when my function is done, the info and next part from head and last are "unable to read memory".
struct node{
int info;
node *next;
};
node *head, *last;
void addelement_end(node *head, node *last, int element)
{if (head == NULL)
{ node *temp = new node;
temp->info = element;
temp->next = NULL;
last = temp;
head = temp;
}
else {node*temp = new node;
last->next = temp;
temp->info = element;
temp->next = NULL;
last = temp;
}
}
void show(node* head, node *last)
{
if (head==NULL)
cout << "Empty list";
else
while (head != NULL)
{
cout << head->info << " ";
head = head->next;
}
}
int _tmain(int argc, _TCHAR* argv[])
{
int x, n, i;
cout << "how many numbers";
cin >> n;
head = last = NULL;
for (i =1; i <= n; i++)
{
cin >> x;
addelement_end(head, last, x);
}
show(head, last);
return 0;
}

Answer 1

It's a very common error. Here is a similar illustration of the problem:
int change_a(int a) {
a = 42;
}
int main() {
int a = 10;
change_a(a);
printf("%d\n", a);
return 0;
}
This will print 10 because in the function change_a you are only modifying a copy of the value contained in the variable a.
The correct solution is passing a pointer (or using a reference since you are using C++).
int change_a(int *a) {
*a = 42;
}
int main() {
int a = 10;
change_a(&a);
printf("%d\n", a);
return 0;
}
But maybe you're going to tell me: "I'm already using a pointer!". Yes, but a pointer is just a variable. If you want to change where the pointer points, you need to pass a pointer to that pointer.
So, try this:
void addelement_end(node **head, node **last, int element)
{
if (*head == NULL)
{ node *temp = new node;
temp->info = element;
temp->next = NULL;
*last = temp;
*head = temp;
}
else {
node *temp = new node;
(*last)->next = temp;
temp->info = element;
temp->next = NULL;
*last = temp;
}
}