CODEWITHSUNDEEP
c++pointerslinked-list
Hans Araya
Hans Araya

Reputation: 345

2 Pointers to same value and changing just one of them - Double linked list C++

I have a class Node with these attributes: 

int value;
Node* next;
Node* prev;

I have a class List with an attribute Node* first when I initialize the list to know which one is the first node.

Assuming I have initialized 2 lists of class List with these values:

list1: 1 | 2 | 3 | NULL
list2: 6 | 7 | 8 | NULL

I want to get to a result where the lists are 'braided':

list1: 1 | 7 | 3 | NULL
list2: 6 | 2 | 8 | NULL

I'll use this method to list1:

void List::test(List list2){
Node* aux1 = first;   //first is 1
Node* aux3 = first;   //first is 1
Node* aux2 = list2.first;   //first of list2 is 6
Node* aux4 = list2.first;  //first of list2 is 6

aux1->next = aux2->next; //aux1->next is now 7 PERFECT
aux1->next->prev = aux1; //aux1->next->prev is now 1 PERFECT

//NOW I WANT TO CHECK aux3->next
cout << aux3->next <<endl; 
//IT PRINTS 7, LIKE AUX1->NEXT, BUT I WANT IT TO BE 2, LIKE IF IT'S BEEN DONE TO THE ORIGINAL LIST1!!
}

How can I make aux3 to stay in there? I thought if I point 2 pointers to just one Node, I could move one of those pointers without touching the attributes of the other pointer.

I know it's a silly problem, but I can't figure it out!

Upvotes: 0

Views: 236

Answers (1)

Kevin DiTraglia
Kevin DiTraglia

Reputation: 26058

Initial

aux1 -->|---|<--|---|
aux3 -->| 1 |-->| 2 |
        |___|   |___|

aux2 -->|---|<--|---|
aux4 -->| 6 |-->| 7 |
        |___|   |___|

after aux1->next = aux2->next

aux1 -->|---|<--- |---|
aux3 -->| 1 |\    | 2 | (no longer anything pointing at 2)
        |___| \   |___|
               \
aux2 -->|---|   \> |---|
aux4 -->| 6 | ---> | 7 |
        |___| <--- |___|

after aux1->next->prev = aux1;

aux1 -->|---|<---  |---|
aux3 -->| 1 |--->  | 7 |
        |___|   /> |___|
               /
aux2 -->|---| /
aux4 -->| 6 |/
        |___|

printing aux3->next

7

EDIT:

OK I think I understand what you are attempting now, you'll want to cache the second node in the first list while you manipulate it:

Node* aux1 = list1.first;   //first of list1 is 1
Node* aux2 = list2.first;   //first of list2 is 6
Node* aux3 = first->next;   //aux3 is 2

aux1->next = aux2->next; //aux1->next is now 7 PERFECT
aux1->next->prev = aux1; //aux1->next->prev is now 1 PERFECT

Now you need to use the cached 2 stored in aux 3 to regain your second list

aux2->next = aux3;
aux3->prev = aux2;

so you should now see:

aux1 -->|---|<--|---|
        | 1 |-->| 7 |
        |___|   |___|

aux2 -->|---|<--|---|
        | 6 |-->| 2 |
        |___|   |___|

Upvotes: 2

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