List with nodes destructor - c++

Before asking my question, I'm aware of this and this questions.
I have a chained List where each node points to the next node. Basically :
class List{
class Node{
Node():next_(NULL){}
~Node(){
if(next_){
delete next_;
next_ = NULL;
}
}
Node* next_;
}
List():head_(NULL),tail_(NULL){}
void add_node(){
tail_->next_ = new Node();
tail_ = tail_->next_;
}
Node* head_;
Node* tail_;
}
I removed all the useless information but the piece of code given above is
enough to get the idea (at least I hope, I'll update it on request).
What concerns me is the destruction part. Assume I have a List where I have
added many node with add_node(). When the List destructor is called, it
recursively calls the destructor of each Node until it reaches the last node
which has next_==NULL. The List is therefore destructed from the end.
I have never had any issue with this until today where I had a list with
~180000 elements (quite a large memory usage but definitely not overwhelming
the RAM) which returns a segfault...
If I do a while loop destroying the last elements (no recursive call of the
destructor) everything works fine.
Is there a limiting number of recursive call of a destructor ? Did I miss
anything else ?

this way recursion can be avoided:
~Node(){
while(next_)
{
Node* temp = next_->next_;
next_->next_ = NULL;
delete next_;
next_ = temp;
}
}

Related

Passing a linked list without memory leak in C++

In many occasions, we need to modify a linked list drastically so we will sometimes create another linked list and pass it to the old one. For example,
struct node { //let's say we have a linked list storing integers
int data;
node* next;
};
and suppose we already have a linked list storing integers 1,2,3.
node* head; //suppose we already store 1,2,3 in this linked list
node* new_head ; //suppose we make a temporary linked list storing 4,5,6,7
head = new_head; //modifying the original linked list
My Question
If I delete head (the old linked list) before the assignment then the whole program will crash.
Conversely, if I do not delete it, then there will be a memory leak.
Therefore, I am looking for a way to modify the linked list without memory leak.
My attempt
I tried to make a helper function similar to strcpy to do my work.
void passing_node(node*& head1, node* head2){ //copy head2 and paste to head1
node* ptr1 = head1;
for (node* ptr2 = head; ptr2 != nullptr; ptr2 = ptr2->next)
{
if (ptr1 == nullptr){
ptr1 = new node;
}
ptr1->data = ptr2->data;
ptr1 = ptr1->next;
}
}
// note that we assume that the linked list head2 is always longer than head1.
However, I still got a crash in the program and I cannot think of any other way to modify this. Any help would be appreciated.
Easier way to avoid memory leak is to avoid raw owning pointers.
You might use std::unique_ptr (or rewrite your own version):
struct node {
int data = 0;
std::unique_ptr<node> next;
};
You can move nodes.
You can no longer copy nodes (with possible double free issue).
so deep_copy might look like:
std::unique_ptr<Node> deep_copy(const Node* node)
{
if (node == nullptr) return nullptr;
auto res = std::make_unique<Node>();
res->data = node->data;
res->next = deep_copy(node->next.get());
return res;
}
I would suggest preallocating the linked list so it's easy to delete every node in one call. The nodes would then just reference somewhere inside this preallocated memory. For example:
struct Node
{
int value;
Node* next;
};
struct LinkedList
{
Node* elements;
Node* first;
Node* last;
Node* free_list;
LinkedList(size_t size)
{
first = nullptr;
last = nullptr;
elements = new Node[size]{0};
free_list = elements;
for (size_t i = 0; i < size-1; ++i)
free_list[i].next = &free_list[i+1];
free_list[count-1].next = nullptr;
}
~LinkedList()
{
delete[] elements;
}
void Add(int value)
{
if (free_list == nullptr)
// Reallocate or raise error.
// Take node from free_list and update free_list to
// point to the next node in its list.
// Update last node to the new node.
// Update the first node if it's the first to be added.
}
void Free(Node* node)
{
// Search for the node and update the previous and
// next's pointers.
// Update first or last if the node is either of them.
// Add the node to the last place in the free_list
}
};
From here you'll have many strategies to add or remove nodes. As long as you make sure to only add nodes to the allocated elements array, you'll never have any memory leak. Before adding, you must check if the array have the capacity to add one more node. If it doesn't, you either have to raise an error, or reallocate a new the LinkedList, copy over all values, and delete the old one.
It becomes a bit more complicated when the array becomes fragmented. You can use a 'free list' to keep track of the deleted nodes. Basically, a LinkedList of all nodes that are deleted.
Just take notice that my code is incomplete. The basic approach is to create an allocator of some sort from which you can allocate a bulk, use segments of it, and then delete in bulk.

Linked List destructor without head?

I'm stuck on a linked list destructor for my class. This is what I have here:
LinkedList::~LinkedList()
{
LinkedList *forward = nullptr;
LinkedList *current = this;
//iterate through list, deleting each element as we go
while (current != nullptr)
{
//set next pointer to current's next
forward = current->next;
delete current; //delete the current memory
current = forward; //reset current to next's pointer
}
}
When I run it, I get a seg fault. I only want to delete just one node from my linked list. Is that possible? Also, I wasn't given a "head" pointer as I was used to from other lists, so I used "this" instead, does that work?
Aka - the .cpp is finding the spot in the linked list to delete, reorganizing the next pointers around it, and then deleting the node (which calls this destructor)
(when I run my program with an empty destructor, it prints out fine, but of course there are memory leaks)
Any help is appreciated!
by definition delete calls LinkedList::~LinkedList, so you have several (in fact infinite) calls to it because of the loop calling delete, so you access to already deleted element with an undefined behavior
just do
LinkedList::~LinkedList()
{
if (next != nullptr)
delete next;
}
or just
LinkedList::~LinkedList()
{
delete next; // delete on nullptr does nothing
}
even personally I prefer to compare to nullptr first
I argee with bruno.
Besides, when you want to delete a node and you don't have the pointer to head, you need a double-linked list.
The double-linked list has two pointers to its prev and next node.
when you need to delete a node, like this:
ListNode* p;
if(p->pre)
p->pre->nxt = p->nxt;
if(p->nxt)
p->nxt->pre = p->pre;
p->nxt = p->pre = NULL;
delete p;

How is this singly linked list destructor causing an infinite loop?

I wrote a singly linked list implementation using a struct. It is not part of an outer class that manages the operations on the list. Instead all of the operations are handled directly with the Nodes.
I understand that if the struct definition was part of a class, say ListManager, calling the destructor on a ListManager instance would just require one to iterate through the linked list managed by the class and delete each Node.
However, since this linked list is not part of an outer class and manages all of the operations itself I am a bit confused as to how to write the destructor.
Version 1 works well, its a recursive call that goes through the list and frees and memory associated with each Node.
Version 2 caused an infinite loop. I don't understand why, as this is one way that I would implement the destructor for a container class that manages the Node linked list.
Version 3 works well but is too verbose.
I ran all three versions using valgrind and python tutor to check for leaks and other issues.
Any help explaining why Version 2 does not work and why it is incorrect to implement the destructor in such a way is appreciated!
Struct Linked List
#include <iostream>
#include <string>
using namespace std;
struct Node
{
int id;
Node* next;
Node(int newId = 0, Node* newNext = NULL)
: id(newId), next(newNext) { }
};
Destructor Version 1
~Node()
{
if (next != NULL)
delete next;
}
Destructor Version 2
~Node()
{
Node* lead = this;
Node* follow = this;
while (follow != NULL)
{
lead = lead->next;
delete follow;
follow = lead;
}
}
Destructor Version 3
~Node()
{
Node* lead = this;
Node* follow = this;
if (follow != NULL)
{
lead = lead->next;
delete follow;
follow = lead;
}
}
Main
int main()
{
Node* head = NULL;
head = new Node(23, head);
head = new Node(54, head);
head = new Node(81, head);
head = new Node(92, head);
delete head;
return 0;
}
In version 2, you have written a loop that clears up the entire list in one destructor call by looping through the list and deleting every element. However, what happens is not that you have just one destructor call. Every time an element is deleted, that calls the destructor again.
So in the end, the delete follow translates to delete this (because follow = this;) for the first invocation. This then causes the destructor of the first node to be called again, causing the endless loop.
The following nodes would be destroyed multiple times, leading to undefined behavior, but it's not even getting there because of that infinite loop.
You only need each Node to delete (at most) one other Node, to eventually delete all the nodes. You re-assigning of local pointers does not affect the structure of the list.
Both 2 and 3 are delete this, which is suspicious at the best of times, plus some irrelevant ceremony, in the destructor. They are both undefined behaviour, deleting the same object (at least) twice.
Your first attempt is close.
Instead of confusing yourself with copying around pointer values, just use an owning pointer type, like std::unique_ptr.
struct Node
{
int id;
std::unique_ptr<Node> next;
Node(int id = 0, std::unique_ptr<Node> next = {})
: id(id), next(std::move(next)) { }
// correct destructor is implicit
};

Deleting forward linked list

There are two known ways (only two ?) to delete forward linked list
One way is the recursive function which is inefficient and would give stack overflow if the list is too large
Another way (the efficient) is a function that iterates and deletes nodes like this:
class Forward_list {
public:
// Constructor...
~Forward_list() { if(head) destroy(); }
void destroy() {
node* prev = nullptr;
while (head) {
prev = head;
head = head->next;
delete prev;
}
}
// functions...
private:
// data members...
node* head;
};
Now what about doing it this way:
class Forward_list {
public:
// Constructor...
~Forward_list() { if(head) delete this->head; }
// functions...
private:
struct node {
~node() { delete this->next; } // <- this way
type data;
node* next;
};
node* head;
// data members...
};
I tested it and it works fine ... I find this way cleaner but not sure if there would be side effects ?
Your solution is technically correct, the only problem I can think of is you can't delete one node without deleting all the nodes that follow.
Since it wasn't mentioned before: when using
delete this->next;
note that the destructor gets invoked recursively, i.e., the approach of deleteing the next node in the destructor is equivalent to the recursive approach of deleteing a list. The recursion is just not as direct and obvious.
~node() { delete this->next; } // <- this way
I find this way cleaner but not sure if there would be side effects ?
Well, the "side effect" will be, that you can't remove any nodes from the list, without deleting the whole rest of the list, since ~node() is called recursively there.
That's probably not what you want to do.

Using the destructor to free linked objects

I have a class called "node". I link a bunch of node objects together to form a linked list. When the "node" destructor is called, it only deletes the first node. How do I iterate through the entire linked list of nodes and delete each node object?
Here is the class definition:
class Node
{
private:
double coeff;
int exponent;
Node *next;
public:
Node(double c, int e, Node *nodeobjectPtr)
{
coeff = c;
exponent = e;
next = nodeobjectPtr;
}
~Node()
{
printf("Node Destroyed");
}
The destructor is called by invoking delete on the pointer to the first node of the linked node list.
Since you don't know how many nodes there are in a list, if you do not have firm bounds on that it's not a good idea to invoke destructors recursively, because each call uses some stack space, and when available stack space is exhausted you get Undefined Behavior, like a crash.
So if you absolutely want to do deallocate following nodes in a node's destructor, then it has to first unlink each node before destroying it.
It can go like this:
Node* unlink( Node*& p )
{
Node* result = p;
p = p->next;
result->next = nullptr;
return result;
}
Node::~Node()
{
while( next != nullptr )
{
delete unlink( next );
}
}
But better, make a List object that has ownership of the nodes in a linked list.
Of course, unless this is for learning purposes or there is a really good reason to roll your own linked list, just use a std::vector (and yes I mean that, not std::list).
How do I iterate through the entire linked list of nodes and delete each node object?
It would be cleaner if you had a separate class to manage the entire list, so that nodes can be simple data structures. Then you just need a simple loop in the list's destructor:
while (head) {
Node * victim = head;
head = victim->next; // Careful: read this before deleting
delete victim;
}
If you really want to delegate list management to the nodes themselves, you'll need to be a bit more careful:
while (next) {
Node * victim = next;
next = victim->next;
victim->next = nullptr; // Careful: avoid recursion
delete victim;
}
Under this scheme, you'll also need to be careful when deleting a node after removing it from the list - again, make sure you reset its pointer so it doesn't delete the rest of the list. That's another reason to favour a separate "list" class.