i'm having some trouble with a function i am trying to write. i cannot delete the second node from a DOUBLY linkedlist if there are just two nodes. here is a snippet of code within my function but i keep getting a segmentation fault, am i doing something wrong because if i comment this part of my code out it works fine:
Code:Node<T> *secondNode = first->next; if(secondNode->next==NULL) //check if there is just 2 nodes. { first->next = secondNode->next; delete secondNode; }
Last edited by Nooby; 04-10-2010 at 10:59 PM.
The only special node is the first one, as you also need to update the head pointer as well.
For every other node, the logic should be the same.
previousNode->next = nodeToBeDeleted->Next; // LL skips the node to be deleted
delete nodeToBeDeleted;
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
I'm thinking in a doubly linked list you would normally have to reset a backward link from node 3 back to node 1, n'est ce pas?
As to the OP, I hope you are first checking that there is only 1 node (as referencing secondNode->next will fail badly if secondNode doesn't exist). Otherwise, presumably either your list isn't quite being built correctly, or other operations are leaving your list in an inconsistent state.
