Is this code safe? (linked list, C++) - c++

The following code seems to work ok to remove a node from a linked list:
bool remove(node * & head, int toBeRemoved)
{
if (head == nullptr) //empty list
return false;
else {
node * temp = head;
//the first node needs to be removed
if (head->data == toBeRemoved) {
head = head->next;
delete temp;
return true;
}
//seek for node and remove it
else {
while (temp->next != nullptr && temp->next->data != toBeRemoved)
temp = temp->next;
if (temp->next->data == toBeRemoved){
node * removeThis = temp->next;
temp->next = temp->next->next;
delete removeThis;
return true;
}
//data to be removed can't be found in the list
else
if (temp->next == nullptr && temp->next->data != toBeRemoved)
return false;
}
}
}
(I understand there's a list implementation in C++ but I'm only trying to understand this algorithm here, not replace it with something else).
Even though the code works when deleting a node placed at the beginning, in between or at the end of the list and I don't see any errors, I still have some doubts about the following line: if (temp->next->data == toBeRemoved).
Since that whole block can be executed when the node to be deleted is the last node (i.e.: temp->next==nullptr) I'm wondering how safe it is to attempt to access temp->next->data.
And even if it's safe, is it a bad programming practice?

No, that line is not safe.
The while loop ends when either temp->next is a null pointer, or temp->next->data contains the value to be removed. If it ends for the first reason, then accessing temp->next->data is invalid, because it's indirecting through a null pointer.
You should change the order of the test after the while loop ends. Check for the null pointer first. If it's not a null pointer, it means you found the element, and you can remove it.
while (temp->next != nullptr && temp->next->data != toBeRemoved) {
temp = temp->next;
}
if (temp->next == nullptr) { // data couldn't be found
return false;
} else { // data was found, remove it
node * removeThis = temp->next;
temp->next = temp->next->next;
delete removeThis;
return true;
}
}

Related

c++ replace values in linked list by changing pointers

Having a problem with linked list. Need to create a method, which will replace data in list, by not creating a new element but by changing pointers. For now I have such method:
void replaceValues(Node* head, int indexOne, int indexTwo)
{
Node* temporaryOne = NULL;
Node* temporaryTwo = NULL;
Node* temp = NULL;
Node* current = head;
int count = 0;
while (current != NULL) {
if (count == indexOne)
{
temporaryOne = current;
}
else if (count == indexTwo)
{
temporaryTwo = current;
}
count++;
current = current->next;
}
current = head;
count = 0;
while (current != NULL) {
if (count == indexOne)
{
head = temporaryTwo;
}
else if (count == indexTwo)
{
head = temporaryOne;
}
count++;
current = current->next;
}
}
I am sure, that exists a more simpler way, how to do it, but I don't fully understand, how it works...
Thanks in advance for help.
I assume that with "replace" you actually mean "swap"/"exchange".
Some issues:
The argument head should be passed by reference, as one of the nodes to swap may actually be that head node, and then head should refer to the other node after the function has done its job.
The node before temporaryOne will need its next pointer to change, so you should stop your loops one step earlier in order to have access to that node and do that.
In some cases head may need to change, but this is certainly not always the case, so doing head = temporaryOne or head = temporaryTwo is certainly not right. In most cases you'll need to link to the swapped node from the preceding node (see previous point).
The next pointer of the node that is swapped will also need to change, as the node that follows it will be a different one than before.
As mentioned already in comments, it is advised to split the task into removals and insertions, as the fiddling with next pointers can get confusing when you try to cover all possible cases, notably making the distinction between the case where the two nodes are adjacent and when they are not.
Here are some functions that split the work into removal, insertion and finally exchanging nodes:
Node* removeNode(Node* &head, int index) {
// If index is out of range, no node is removed, and function returns nullptr
// Otherwise the extracted node is returned.
if (head == nullptr || index < 0) return nullptr;
Node* current = head;
if (index == 0) {
head = head->next;
current->next = nullptr;
return current;
}
while (--index > 0) {
current = current->next;
if (current == nullptr) return nullptr;
}
Node* temp = current->next;
if (temp != nullptr) {
current->next = temp->next;
temp->next = nullptr;
}
return temp;
}
void insertNode(Node* &head, Node* node, int index) {
// If index is too large, node is inserted at the end of the list
// If index is negative, node is inserted at the head of the list
if (index <= 0 || head == nullptr) {
node->next = head;
head = node;
return;
}
Node* current = head;
while (--index > 0 && current->next != nullptr) {
current = current->next;
}
node->next = current->next;
current->next = node;
}
bool exchangeNodes(Node* &head, int indexOne, int indexTwo)
{
// Returns true when successful, false when at least one index
// was out of range, or the two indexes were the same
if (head == NULL || head->next == NULL || indexOne == indexTwo || indexOne < 0) return false;
// To ensure the right order of operations, require the first index is the lesser:
if (indexOne > indexTwo) return exchangeNodes(head, indexTwo, indexOne);
Node* two = removeNode(head, indexTwo);
if (two == nullptr) return false; // out of range
Node* one = removeNode(head, indexOne);
insertNode(head, two, indexOne);
insertNode(head, one, indexTwo);
return true;
}

Deleting the root node of a BST

I'm trying to make a remove function (without the use of recursion), that passes in the value of the node I want to remove.
My current problem is that my destructor seems to be making the program crash (getting a runtime error), which is probably because I'm not deleting the root node properly in my remove function.
The code I'm trying to use to remove the root is here:
bool BST::remove_root (int val)
{
if (val == root_->val)
{
if (root_->left == NULL && root_->right != NULL)
{
Node* temp = root_->right;
delete root_;
root_ = NULL;
size_--;
root_ = temp;
return true;
}
else if (root_->right == NULL && root_->left != NULL)
{
Node* temp = root_->left;
delete root_;
root_ = NULL;
size_--;
root_ = temp;
return true;
}
else
{
Node *curr = root_->right, *child = root_->left;
delete root_;
root_ = NULL;
size_--;
root_ = curr;
Node* temp = curr;
while (temp->left != NULL)
temp = temp->left;
temp->left = child;
return true;
}
}
}
In my code, just using val by itself is the value being passed in to the remove function (the one I want to remove). When I dereference things like root_->val, I'm accessing the val in my BST class.
I don't really understand why this wouldn't be able to continuously delete the root node (like I said I think it's the destructor making the program crash), but I feel like it could be invalid pointers?

Return inside loop for C++ breaks loop?

I am reversing a doubly linked list. My function for the same is :
Node* Reverse(Node* head)
{
// Complete this function
// Do not write the main method.
Node* temp = new Node();
if ( head == NULL) { return head; }
while ( head != NULL) {
temp = head->next;
head->next = head->prev;
head->prev = temp;
if (temp == NULL ) { break; }
head = temp;
}
return head;
}
This works correctly.
Instead of using the break command, if I do 'return head' than the function exits the while loop and has a compile error: control reaches end of non-void function [-Werror=return-type]
Node* Reverse(Node* head)
{
// Complete this function
// Do not write the main method.
Node* temp = new Node();
if ( head == NULL) { return head; }
while ( head != NULL) {
temp = head->next;
head->next = head->prev;
head->prev = temp;
if (temp == NULL ) { return head; }
head = temp;
}
}
What could be the reason behind this?
The reason is that the compiler doesn't know that temp will always end up being NULL at some point, it only knows the loop can end, and the function will have no return statement.
As CompuChip pointed out, you may add the following line at the end, to appease the compiler:
throw std::runtime_error("Control should never reach this point");
Or you may simply return NULL at the end.
What you know, but the compiler doesn't, is that the condition temp == NULL will be true before head != NULL becomes false. It assumes that the while loop may end naturally (with head == NULL) and therefore a return statement is missing.
Changing the while loop condition to 1 (true) will probably resolve the warning. (It is a warning after all, not really an error; it becomes an error because you've asked it to using -Werror).
Better still would be to avoid changing altering control flow as much as you do (avoid break where possible). This makes the code easier to understand for other humans and for the compiler:
Node* Reverse(Node* head)
{
if (head == NULL) { return head; }
Node* temp = head->next;
while (temp != NULL) {
head->next = head->prev;
head->prev = temp;
head = temp;
temp = head->next;
}
return head;
}

Deletion in a linked list

Qn) Given only a pointer to a node to be deleted in a singly linked list, how do
you delete it?
I am trying to delete the last element i.e., 1 but the else part goes into an infinite
loop printing garbage values.
Original link.
int main()
{
struct Node* head = NULL;
push(&head, 1);
push(&head, 4);
push(&head, 6);
push(&head, 8);
print(head);
del_p(head->next->next->next);
cout << endl;
print(head);
return 0;
}
void del_p(struct Node* current)
{
struct Node* temp;
if (current->next != NULL)
{
temp = current->next;
current->data = temp->data;
current->next = temp->next;
free(temp);
}
else
{
free(current);
current = NULL;
}
}
The else branch of your function tries to reassign current to NULL. This is problematic because current is a local copy of the pointer passed in. That is, you can't modify the value of the original pointer.
This is why you are receiving garbage, because you're accessing a node whose memory has already been deallocated.
You either need a double pointer, or preferably a reference to the node:
void del_p(struct Node*& current)
If you pass in the node to be deleted and the head node then you can loop until you find the node prior to the node to be deleted. You then need to point the prior node to the node that is pointed to by the node to be deleted and then you can free the node you want to delete.
void delete(struct Node* to_delete, struct Node* head)
{
// check if node to be deleted is the head
if (to_delete == head)
{
head = to_delete->next;
return;
}
// make a local copy of the head just in case as to not alter it
struct Node* tempHead = head;
while(tempHead->next != to_delete)
{
tempHead = tempHead->next;
}
tempHead->next = to_delete->next;
free(to_delete);
}
Just as a disclaimer I haven't tested this code, but conceptually it should work.
The typical algorithm for deleting a node on a linked list would follow the next steps:
Get a temp pointer started in Head.
Move your temp to the node you want to delete (in this case one before the last: temp->next == NULL).
Free the memory for temp2.
Set the pointer of temp->next to NULL.
Return the pointer to head.
Now this is not the only algorithm, there are a lot of ways you can accomplish this. The following code would be my solution to the function del_p (if you would want to delete the last node):
void del_p(struct Node *head)
{
if (head != NULL)
{
struct Node *temp = head;
while (temp->next != NULL) temp = temp->next;
free(temp);
}
}
You can make this code a little more general to make it possible to delete any Node, by passing a pointer to that node (or a value), the code would look as follows:
void del_p(struct Node **head, struct Node *delete_node)
{
if (head != NULL)
{
struct Node *temp = *head;
if (temp == delete_node)
{
*head = (*head)->next;
free(temp);
}
else
{
while (temp->next != NULL && temp->next != delete_node)
temp = temp->next;
if (temp->next != NULL && delete_node != NULL)
{
temp->next = delete_node->next;
free(delete_node);
}
}
}
}
Hope this works for you, this code isn't tested, but tell me if you have troubles!

Delete a node from the middle of a C++ queue

I have a linked list with a c-style ctor and dtor.
I just got too frustrated when this if statement decided not to test true, putting me in
an infinite loop. I dont understand why it will never test true.
I am trying to delete a node (the address of a class object) from my LinkedList.
Maybe someone could help me out?
Node *Current = first_; // I want this to be my only Node Ptr Varaible Declaration.
if ( NULL == first_ )
std::cout << "Cannot delete from an empty list: \n";
while ( Current != NULL )
{
if ( first_->data_ == node->data_ )
{
//check to see if we are deleteing the head.
first_ = first_->next_;
--listLen_;
delete Current;
std::cout << "Head Deleted!\n";
}
if ( Current->data_ == node->data_ ) // FOR SOME REASON this is never true?
{
--listLen_;
node->data_ = NULL;
Current = Current->next_;
node->data_ = Current;
}
else // we must not of found it. // else should match previous i
{
Current->prev_ = Current;// since we are not deleting the first node we are OK here.
Current = first_->next_;
if ( Current->next_ == NULL ) // see if we are at the end of the list.
{
first_ = NULL;
last_ = Current->prev_;
}
}
}
return;
This should really be rewritten, since it has too many problems...also why not use a STL container? I assume this is a homework question.
The answer to the infinite loop is the else case that increments to the next node:
Current = first_->next_;
This will make you loop forever if the data is not found in in the first two nodes...since you will set the next test to the first's next node always and it will never set the current to NULL provided there are more than 2 nodes in the list.
Keep your loops small, it easier to figure out what went wrong. Assuming your data compare makes sense, look at this the following:
curr = first_;
while( curr && (curr->data_ != node->data_) ) {
curr = curr->next_;
}
if (!curr) return // didnt find it, nothing to remove
if ( curr == first_ )
first_ = curr->next_
else
curr->prev_->next_ = curr->next_
curr->next_->prev_ = curr->prev_ // always fix next's prev
delete curr
I'm not entirely sure what you're trying to accomplish, but I'm certain you're doing it wrong. If you're merely trying to remove an element from a doubly-linked list that matches node->data_, it's as easy as this:
Node *Current = first_;
while (Current != NULL)
{
if (Current->data_ == node->_data)
{
//If Current isn't the head of the list, set prev to next
if (Current != first_)
Current->prev_->next_ = Current->next_
else
{
first_ = Current->next_;
if (first_ != NULL)
first_->prev_ = NULL;
}
//If Current isn't the tail of the list, set next to prev
if (Current->next_ != NULL)
Current->next_->prev_ = Current->prev_
else if (Current->prev_ != NULL)
Current->prev_->next_ = NULL;
delete Current;
Current = NULL;
}
else
{
Current = Current->next_;
}
}
return;
You don't show where node comes from or how data_ is defined, but if it is a pointer type, you probably need to compare the contents, not the addresses.
Assuming that data_ is a pointer to something and what it points to has operator== defined or is a built in type and it has the value you are looking for then you can do this instead:
if ( *Current->data_ == *node->data_ )
if first_->data_ == node->data_ ever evaluates to true then the second if statement will always evaluates to true then the second if condition will always evaluate to false Current->data_ == node->data_ because on the first iteration first_ == Current and you delete Current without ever updating it
To delete a node from a linked list you seem to be doing way too much work.
Not really an answer here for the actual question but a suggestion.I would never iterate through a linked list to delete an entry in the list. Each entry should have a valid next and previous pointer and when you go to delete an entry from the list you just make the previous record point to the next one and vice versa to remove yourself from the list. An empty list should have a head record and a tail record that just point to each other and all valid entries are inserted in between.
Delete a node with value passed.
void deleteBegin()
{
Node* temp =Head;
if(temp==NULL)
return;
Head=Head->next;
free(temp);
}
void deleteMiddle(int _data)
{
Node* curr = Head;
Node* prev = Head;
if(curr==NULL)
return;
if(curr->next==NULL)
{
deleteBegin();
return;
}
while(curr->next!=NULL && curr->data!=_data)
{
prev=curr;
curr=curr->next;
}
if(curr->data == _data)
{
if(prev==curr)
{
deleteBegin();
return;
}
prev->next = curr->next;
free(curr);
}
else
{
cout<<"Element Not Found\n";
return;
}
}