How do you move between nodes of a linked list? - c++

Question

This is a piece of code that tries to build a linked list.
struct node {
char name[20];
int age;
int height;
node* next; // Pointer to the next node
};
node* startPTR = NULL;
void addNode_AT_END() {
node *temp1;
node *temp2;
temp1 = new node;
cout << "Enter the name : ";
cin >> temp1->name;
cout << endl << "Enter the age : ";
cin >> temp1->age;
cout << endl << "Enter height : ";
cin >> temp1->height;
temp1->next = NULL;
if( startPTR == NULL) {
startPTR = temp1;
} else {
temp2 = startPTR;
while( temp2->next != NULL )
temp2 = temp2->next;
temp2->next = temp1;
}
}
The following is diagram after 2 back to back calls to the above function.
start = addr1;
|
V
(addr1) ----> (addr2) ----> (NULL) at end
^
|
temp2
where addr1 and addr2 are the address of the first and second nodes respectively.
What happens after the third call ? How the iteration will go on for the third call? I am unable to understand how the list links up after the second call.According to me all that has been build up till know will vanish.Then how will list move further ? How is node placed during the third call?

Answer 1

Here is where all the magic happens:
1. temp2 = startPTR;
2. while( temp2->next != NULL )
3. temp2 = temp2->next;
4. temp2->next = temp1;
First, temp2 will point to the beginning of the list. In lines 2 and 3, you change temp2 to the next node until you reach the node where temp2->next is NULL. This node is the last node of the list, regardless of the size of the list.
Finally, in line 4 you change temp2->next to temp1 so now it points to the new node (that is last node now points to the new node). temp1->next is also NULL, so temp1 now represents the end of the list.
After line 1 you have
start = addr1;
|
V
(addr1) ----> (addr2) ----> (NULL)
^
|
temp2
temp2->next is not NULL (it is addr2), so you iterate and execute line 3 and you get:
start = addr1;
|
V
(addr1) ----> (addr2) ----> (NULL)
^
|
temp2
temp2->next is now NULL. So you stop the loop and execute line 4 and you get:
start = addr1;
|
V
(addr1) ----> (addr2) ----> (addr3) ----> (NULL)
^ ^
| |
temp2 temp1
Note: Do you know how pointers work? Imagine this: You have a node, which is some data in the memory. When you have variables in memory, these variables have addresses. Let's say addr1 is 10, addr2 is 150 and addr3 (which is the node just newed) is 60. start has value 10. Therefore, "pointing" to the first node of the list (that is using this address, you have access to its data). One of these data, is the next field. The first node's next field has value 150, thus pointing to the next node. When you say temp2 = start, you put number 10 in temp2, at this point temp2->next has value 150. When you say temp2=temp2->next, you simply put value 150 in temp2, overwriting the previous value. This way you have effectively moved your pointer from pointing to the first node, to now pointing to the second node. Now temp2->next is NULL (that is 0). When you now say temp2->next=temp1, you put value 60 in the next field of temp2. So now temp2->next is 60. temp2->next->next is NULL.

Answer 2

It's pretty simple. The while cycle will move temp2 to the last element. Then the node you created, pointed by temp1, is assigned as temp2's next node

Answer 3

I don't get what's bothering you. During any call while() loop will go through all the nodes in the list untill it reaches the end, and then set the pointer in the last one to the newly allocated node (temp1).
temp1 and temp2 are pointers. they do not store data of the node, they store address in memory where data is stored. so at the end of first iteration, after startPTR = temp1 startPTR points to the same address that temp1 pointed to. it doesn't matter if temp1 is still there, since now startPTR points to the node. at the end of the second call temp2->next=temp1 (temp2==startPTR at this moment) makes next field of the node point to the newly allocated temp1