Double free in c++ destructor - c++

I'm trying to implement a linked list in C++. The list contains a pointer to a node type allocated on the heap
The code is as follow:
#include <memory>
template<typename T>
class node {
public:
node(T v) : value(v) {}
~node() = default;
T value;
node *next;
};
template<typename T, class Allocator = std::allocator<node<T>>>
class linked_list {
private:
node<T>* head;
Allocator alloc;
public:
linked_list() : head(nullptr) {}
~linked_list() {
for (auto start = head; start != nullptr; start = start->next) {
start->~node();
alloc.deallocate(start, 1);
}
}
void push_back(T value) {
node<T> *new_node = alloc.allocate(1);
new (new_node) node<T>(value);
if (head == nullptr) {
head = new_node;
return;
}
head->next = new_node;
head = new_node;
}
};
In main.cpp:
#include "linked_list.hpp"
int main() {
linked_list<int> a;
a.push_back(4);
a.push_back(5);
return 0;
}
When I ran it I got double free detected in cache T2.
What did I do wrong with the destructor ?

This is a common newbie error. You modified your loop control variable.
for (auto start = head; start != nullptr; start = start->next)
{
start->~node();
alloc.deallocate(start, 1);
}
You modified start (deleting the memory) in the for loop's body and then tried to dereference the pointer you just deleted in its continuation expression. BOOM! You are fortunate that the runtime library was smart enough to catch this and give you the "double free" error rather than, say, launch a nuclear missile.
This is one place a while loop is better than a for loop.
while (head)
{
auto to_del = head;
head = head->next;
to_del ->~node();
alloc.deallocate(to_del, 1);
}
I've left out a lot of commentary about your antiquated techniques because they don't relate to the problem you're having, but if you really want to substitute in a different kind of allocator you should look into using allocator_traits for allocation, construction, destruction, and deallocation of your elements.
There are other problems, such as push_back being completely wrong as to where it inserts the new node. Replacing head->next = new_node; with new_node->next = head; will at least keep your program from orphaning all of the new nodes.

Related

Linked Lists Append To Tail

I am trying to create an appendToTail function which will add a node to the end of a singly linked list.
I am having trouble in adding a node if the head is NULL(the linked list is empty)
class Node {
private:
Node* next;
int data;
public:
Node(int d, Node* n = NULL)
: data(d)
, next(n)
{
}
void appendToTail(int);
//other trivial functions(getters and setters etc. ) defined and
//declared
};
void Node::appendToTail(int d)
{
Node* end = new Node(d);
Node* n = this;
if (n == NULL)
n = end;
else {
while (n->next != NULL)
n = n->next;
n->next = end;
n->next->next = NULL;
}
end = NULL;
delete end;
}
int main()
{
Node* n = NULL;
n->appendToTail(5);
std::cout << n->getData(); //getData() is a function which
//retrieves the Data member variable
}
I am expecting to get 5 but I am getting an error which appears to be caused because my node remains null.
Now with modern C++ idioms we use smart pointers instead of raw pointers, it gives you the benefit of RAII (Resource acquisition is initialization) mechanism. In addition if you want an elegant solution to your problem you should introduce a List class with which you can express more clearly the concept of an empty list. It would give something like this:
#include <memory>
#include <iostream>
class List
{
public:
class Node
{
private:
std::shared_ptr<Node> next;
int data;
public:
Node(int d):next(nullptr),data(d){}
inline int getData() const {return data;}
inline std::shared_ptr<Node> getNext() const {return next;}
friend List;
};
List():head(nullptr),tail(nullptr){}
void appendToTail(int );
inline std::shared_ptr<Node> getHead() const {return head;}
inline std::shared_ptr<Node> getTail() const {return tail;}
private:
std::shared_ptr<Node> head;
std::shared_ptr<Node> tail;
};
void List::appendToTail(int d)
{
auto newTail = std::make_shared<Node>(d);
if (head == nullptr)
{
head = tail = newTail;
}
else
{
tail->next = newTail;
tail = newTail;
}
}
int main()
{
List l;
l.appendToTail(5);
std::cout<<l.getHead()->getData();
}
But you should definitely prefer std::list<T> or std::vector<T>.
Unfortunately there several errors with your approach. Semantic errors and a logical error with your interpretation of a linked list. Let's start with your initial misunderstanding. You cannot add a new tail to an empty list. Because it is emtpy. Meaning, not yet existing. Only if some object is existing/instantiated you can add a tail. You cannot add something to not existing stuff. So your idea to start with a Node* n = nullptr cannot work logically.
Additionally you are dereferencing a nullptr (major bug). That is also the main problem of your code. Nothing works. You need an instantiated object, before you can call it's member functions.
So before you can populate the list, you need to create/instantiate it initially. So you need to explicitly create the first node in your main function with
Node* n = new Node (5)
Then the list is existing and from now on you can add new members with calling appendToTail.
There are more semantic errors in your code which have luckily no side effects.
You must not delete the 'end' variable in your function. You want to keep the newly allocated memory for the new tail. But you introduced an additional sematic error by setting 'end' to nullptr and then call delete. Deleting a nullptr is a noOp and will do nothing. So, although you have a semantic error, this will not cause any trouble.
There is more:
For a pointer to Null you should always use nullptr.
And, your
if (n == NULL)
is always false. Before that, you assigned this to n. This is never NULL. You can delete the if else. Keep the statements from the else, except the
n->next->next = NULL;
That's not necessary. The constructor did that already for you. As explained, the next 2 statements should also be elimanted.
Additionally you may want to read a little more on the concept of linked lists.
I hope I could help a little

Nodes added to linked list revert back to NULL after function

I'm reading lines from a file and storing them into linked list.
void add(llist *list, somevalue) {
Node *newnode = (Node *) malloc(sizeof(Node));
newnode->value = somevalue;
newnode->next = list->head;
list->head = newnode;
}
and I call this function from an initialize function which opens the file and reads lines from the file.
void init() {
llist *list = (llist *) malloc(sizeof(llist));
//
//bunch of file i/o codes
//
while (read file until it returns NULL) {
add(list, line);
//if I try to print the values of the list here it works.
}
//Outside the loop, the head is back to NULL
}
And another problem that I realized is the values get concatenated every time I try to print the value. That is to say, the output would be:
First Loop: Tony
Second Loop: Peter
TonyPeter
Third Loop: Mir
PeterMir
TonyPeterMir
How do I fix it so the add function permanently adds the node to the linked list?
Why would the values be jumbled up like that?
----EDITED----
The list is a global variable, and here are some more snippets from the init function. This is the while loop with the problem:
//open file
//initialize & declare pointers
while (1) {
for (i = 0; i < max; i++) {
value[i] = '\0';
}
if (!(fgets(value,max,f))) {
//segfaults if I try to print out the list inside this block.
break;
}
add(list, value);
//the values are well separated in this statement
printf("id is %s\n", list->head->value);
//This print list works, but works weird as shown above.
print_list(list);
}
fclose(f);
//This print list doesn't work, the list is NULL
print_list(list);
And this is the print list function:
void print_users(llist *list) {
ListNode *e;
if (list->head == NULL) {
printf("NO USERS\r\n");
return;
}
e = list->head;
while (e != NULL) {
puts(e->id);
e = e->next;
}
}
So I don't have a good grasp at all on what you're exactly trying to do here, so we can only do but so much. You may consider posting a MCVE. However, I may be able to give you some pointers on building a linked list. I directly copied your add function into a linked list class that I just hurriedly built, and it worked fine, so there may be something else in your llist class that is causing the issue, or it could be something else in your code. The class and a brief description of the class are listed below.
basic node class
Note: I used templates, but you could just as easily remove the template statements and replace T with any type.
template<typename T>
class node {
private:
T data;
node* next;
public:
// The C++11 rule of 5
// Default Constructor
node(T value = T(), node* nxt = nullptr) : data(value), next(nxt) { }
// Copy Constructor
node(const node<T>& src) : data(src.data), next(src.next) { }
// Move Constructor
node(node<T>&& src) : data(src.data), next(src.next) {
src.data = T();
src.next = nullptr;
}
// Copy Assignment Operator
node<T>& operator=(const node<T>& src) {
this->data = src.data;
this->next = src.next;
return(*this);
}
// Move Assignment Operator
node<T>& operator=(node<T>&& src) {
this->data = src.data;
this->next = src.next;
src.data = T();
src.next = nullptr;
}
// Destructor
~node() {};
// Some functions to help with encapsulation
void set_next(node* nxt) {
this->next = nxt;
}
void set_data(T value) {
this->data = value;
}
node* get_next() {
return(this->next);
}
T& get_data() {
return(data);
}
};
linked list class body
Since you're using dynamic memory, you need to make sure you adhere to the rule of 3 or 5 depending on whether or not you're using C++11.
template<typename T>
class llist {
private:
node<T>* head;
public:
llist();
llist(const llist& src);
llist(llist&& src);
llist(~llist);
llist& operator=(const llist& src);
llist& operator=(llist&& src);
void push();
void insert();
};
default constructor
Nothing fancy here.
template<typename T>
llist<T>::llist() : head(nullptr) { }
copy constructor
Since you're using dynamic memory this is crucial
template<typename T>
llist<T>::llist(const llist& src) {
node<T>* tmp = src.head;
while(tmp) {
this->push(tmp->get_data());
}
}
move constructor
template<typename T>
llist<T>::llist(llist&& src) {
// delete current nodes
node<T>* tmp = this->head;
while(tmp) {
tmp = head->get_next();
delete head;
head = tmp;
}
// steal the sources list
this->head = src.head;
src.head = nullptr;
}
destructor
template<typename T>
llist<T>::~llist() {
node<T>* tmp = this->head;
while(tmp) {
tmp = head->get_next();
delete head;
head = tmp;
}
}
copy assignment operator
template<typename T>
llist& llist<T>::operator=(const llist<T>& src) {
node<T>* tmp = src.head;
while(tmp) {
this->push(tmp->get_data());
}
return(*this);
}
move assignment operator
template<typename T>
llist& llist<T>::operator=(llist<T>&& src) {
node<T>* tmp = this->head;
while(tmp) {
tmp = head->get_next();
delete head;
head = tmp;
}
this->head = src.head;
src.head = nullptr;
return(*this);
}
push member
this is essentially opposite of your add member.
template<typename T>
void llist<T>push(T data) {
node<T>* new_node = new node<T>(data);
if(this->head) {
node<T>* tmp = this->head;
while(tmp->get_next()) {
tmp = tmp->get_next();
}
tmp->set_next(new_node);
} else {
this->head = new_node;
}
}
insert member
This is essentially your add member.
template<typename T>
void llist<T>insert(T data) {
node<T>* new_node = new node<T>(data, this->head);
this->head = new_node;
}
I don't know if this will help, and you probably already have and know most of this, but I hope it helps.
In this code, it would appear that you attempted to 'malloc' space for a "llist" user defined object.
void init() {
llist *list = (llist *) malloc(sizeof(llist));
//
//bunch of file i/o codes
//
while (read file until it returns NULL) {
add(list, line);
//if I try to print the values of the list here it works.
}
//Outside the loop, the head is back to NULL
}
First, you tagged this as C++. In C++, you simply must use new and delete. The C++ compiler does not associate "malloc" with the ctor / dtor of your user created object called "llist". And I assure you that you really do want to create these two methods, even when each are simple. Really.
On the other hand, the C++ compiler does provide New and Delete, and will automatically invoke the ctor and dtor when appropriate for both dynamic variables (in heap), and automatic variables (on stack). The compiler will not support this with malloc.
Second, your function init() does not return or otherwise deliver the value of the automatic variable you named "list" to any other scope. (Typically, a list lifetime exceeds the life of any function that uses or creates it.)
So your object "list" only exists within the scope of the function init(), within the lifetime of init(). Not very useful.
So the handle to the list of 'malloc'ed things is lost, no longer accessible to anything else. After init(), where did you plan for the listHead to reside?
Even if you used new (and delete) the code still does not deliver the listHead anywhere.
To further your program, you need perhaps 1 of 2 things:
1) a return (from the function) of the "list" handle (I've been calling it "listHead" as you intended, right?)
llist* init() {
llist *listHead = ...
return(listHead);
}
OR
2) a parameter reference which your init function changes. This places the list head outside of init().
void init( llist** listHead) {
llist *list = ...
*listHead = list;
}
You might look into, and take hints from std::list, which has 40+ methods, though you might only need 10. For the methods you plan to implement, you should strive to conform to and use similar names and parameters.
Perhaps you meant to use a class data attribute with the label list (it is quite difficult to imagine this from what you provided). In this case, you should distinguish data attributes names to help you remember what it is, and that it has a different scope. For instance, I would use m_listHead. The prefix m_ (or often, simply the one char prefix '_') simply indicates to the reader that this symbol is a data attribute of a class. This idea is a common c++ idiom, and not enforced by the compiler, but is often part of a coding-standard.
Good luck.

How to assign an instance of an object to a pointer that was null?

Code:
#include <iostream>
using namespace std;
class Node {
public:
Node *next;
int value;
Node(int value) {
this->next = nullptr;
this->value = value;
}
};
class LinkedList {
private:
Node *head;
Node *tail;
public:
LinkedList() {
this->head = nullptr;
this->tail = nullptr;
}
void addToEnd(int value) {
if(head == nullptr)
this->head = new Node(value);
else
this->tail->next = new Node(value);
this->tail = this->tail->next;
}
void print() {
for(Node *n = this->head; n != nullptr; n = n->next)
cout<<n->value<<" ";
cout<<endl;
}
};
int main() {
LinkedList *list = new LinkedList();
list->addToEnd(21);
list->addToEnd(25);
list->addToEnd(56);
list->addToEnd(24);
list->print();
return 0;
}
My problem is, when I am assigning an instance of Node to this->head, the program crashes. Is there different way of assigning an instance to a pointer that was initially nullptr?
This code structure works fine on Java, I came from Java, that is why I have difficulty on C++'s pointers.
EDIT
I pasted the right code now, I'm sure. Sorry.
Ok, I have solved the problem. So, the problem is not about allocating an object to a class member, but, the problem is accessing a nullptr member: this->tail.
I edited this method, and the program now runs the way I wanted.
void addToEnd(int value) {
Node *n = new Node(value);
if(head == nullptr)
this->head = n;
else
this->tail->next = n;
this->tail = n;
}
Thanks for your help people, this question is now SOLVED. :)
I don't know about "it crashes", but the following line is not valid:
this->head = Node(value);
head is a pointer-to-Node but you're trying to assign a Node to it. Even if this automatically took the address of the temporary you created on the RHS (which it doesn't), you'd have a pointer to a local variable that doesn't exist for very long.
You should be getting a compilation error for that.
You'd have to use new to create a new object dynamically — be sure to write code to free that memory later!
You're similarly messing up dynamic memory allocation in main, where you have a needless memory leak. LinkedList list; will do fine, there.
You need to allocate memory for your Node instances. The quickest way is to call new Node(value) wherever you call Node(value). However if I were you I'd consider using shared_ptr<Node> rather than plain pointers.

List destructor in C++

I've just implemented the Linked List. It works perfectly fine but even tough I've seen notation I am unable to create working destructor on Node, that's why it's unimplemented here in code.
I need to implement working destructor on node
Destructor of List but this one is simple I will just use the destructor from Node class(but I need this one).
Make the List friendly to Node so I will not have to use getNext(), but I think I can
handle it myself(not sure how, but I'll find out).
Please look at the code it is perfectly fine, just will work if you copy it.
#include <cstdio>
#include <cmath>
#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
class Node {
public:
Node(Node* next, int wrt) {
this->next = next;
this->wrt = wrt;
}
Node(const Node& obiekt) {
this->wrt = obiekt.wrt;
this->next = obiekt.next;
}
~Node() {}
void show() {
cout << this->wrt << endl;
}
int getWrt(){
return this->wrt;
}
Node* getNext(){
return this->next;
}
void setNext(Node* node){
this->next = node;
}
private:
Node* next;
int wrt;
};
class List{
public:
List(int wrt){
this->root = new Node(NULL, wrt);
}
List(const List& obiekt){
memcpy(&this->root,&obiekt.root,sizeof(int));
Node* el = obiekt.root->getNext();
Node* curr = this->root;
Node* next;
while(el != NULL){
memcpy(&next,&el,sizeof(int));
curr->setNext(next);
curr = next;
next = curr->getNext();
el = el->getNext();
/* curr->show();
next->show();
el->show(); */
}
}
void add(int wrt){
Node* node = new Node(NULL, wrt);
Node* el = this->root;
while(el->getNext() != NULL){
//el->show();
el = el->getNext();
}
el->setNext(node);
}
void remove(int index){
Node* el = this->root;
if(index == 0){
//deleting old one
this->root = this->root->getNext();
}
else{
int i = 0;
while(el != NULL && i < index - 1){
// el->show();
el = el->getNext();
i++;
}
if(el!=NULL){
Node* toRem = el->getNext();
Node* newNext = toRem->getNext();
el->setNext(newNext);
//deleteing old one
}
}
}
void show(){
Node* el = this->root;
while(el != NULL){
el->show();
el = el->getNext();
}
}
~List(){}
private:
Node* root;
};
int main(){
List* l = new List(1); //first list
l->add(2);
l->add(3);
l->show();
cout << endl;
List* lala = new List(*l); //lala is second list created by copy cosntructor
lala->show();
cout << endl;
lala->add(4);
lala->remove(0);
lala->show();
return 0;
}
I suggest you to start with implementing destructor of List. Since you dynamically allocated memory by using new, you should free it by using delete. (If you used new[], it would be delete[]):
~List()
{
Node* currentNode = this->root; // initialize current node to root
while (currentNode)
{
Node* nextNode = currentNode->getNext(); // get next node
delete currentNode; // delete current
currentNode = nextNode; // set current to "old" next
}
}
Once you have proper destructor, you should try whether your copy constructor is correct:
List* lala = new List(*l);
delete l; // delete list that was used to create copy, shouldn't affect copy
you will find out that your copy constructor is wrong and also causes your application to crash. Why? Because purpose of copy constructor is to create a new object as a copy of an existing object. Your copy constructor just copies pointers assuming sizeof(Node*) equal to sizeof(int). It should look like this:
List(const List& list)
{
// if empty list is being copied:
if (!list.root)
{
this->root = NULL;
return;
}
// create new root:
this->root = new Node(NULL, list.root->getWrt());
Node* list_currentNode = list.root;
Node* this_currentNode = this->root;
while (list_currentNode->getNext())
{
// create new successor:
Node* newNode = new Node(NULL, list_currentNode->getNext()->getWrt());
this_currentNode->setNext(newNode);
this_currentNode = this_currentNode->getNext();
list_currentNode = list_currentNode->getNext();
}
}
Also your function remove is wrong since it "removes" reference to some Node but never frees memory where this Node resides. delete should be called in order to free this memory.
"I need to implement working destructor on node" - No, you don't. Node itself doesn't allocate any memory, thus it shouldn't free any memory. Node shouldn't be responsible for destruction of Node* next nor cleaning memory where it's stored. Don't implement destructor nor copy constructor of Node. You also want to read this: What is The Rule of Three?
"Make the List friendly to Node so I will not have to use getNext()" - You want to say within Node class, that class List is its friend:
class Node
{
friend class List; // <-- that's it
Note that from these 5 headers that you include your code requires only one: <iostream>.
Also note that writing using namespace std; at the beginning of the file is considered bad practice since it may cause names of some of your types become ambiguous. Use it wisely within small scopes or use std:: prefix instead.
The linked list destructor will be called either when delete is used with a previously allocated pointer to a linked list or when a linked list variable goes out of scope (e.g., a local variable is destroyed when returning from a function).
The destructor for the linked list should be responsible to free the memory you previously reserved for the nodes (i.e., using add operation). So, basically, you need to traverse the list of nodes and apply the delete operation on each one of them. There is a little trick: when you are about to delete a node you must be careful not to lose the pointer to the next element (when a node is deleted you cannot be sure that next member will still be valid).
If you want to create a destructor for your Node, it should be quite simple actually.
Here it is:
class Node {
private:
int wrt;
Node* next;
public:
Node(Node* next, int wrt) {
this->next = next;
this->wrt = wrt;
}
// Your desired destructor using recursion
~Node() {
if ( next != NULL )
delete next;
}
};
It's that simple :)
Basically, right before the Node is deleted, if next is not empty, we delete next, which will again call the destructor of next, and if next->next is not empty, again the destructor gets called over and over.
Then in the end all Nodes get deleted.
The recursion takes care of the whole thing :)

How do you allocate memory for an linked list when passing its reference instead of its pointer?

How do you allocate memory for an link list when passing its reference instead of its pointer?
For example:
struct node {
string info;
node *next;
};
void add(node &aNode){
//if I use
node *newNode;
newNode = new node;
aNode.next = newNode; //aNode.next = newNode; doesn't work either
//allocating on heap seems to give segmentation error.
}
int main() {
node *aNode;
aNode = new node;
add (aNode);
}
Compiler error: error: invalid initialization of reference of type ‘node&’ from expr
alternatively if I use
int main() {
node aNode;
add (aNode);
add (aNode);
aNode.next->next->info = "abc";
string a = aNode.next->next->info;
}
This give segmentation fault.
So is it possible to allocate for an linked list just with its reference? (this is C++)
It should be
node * newNode = new node;
aNode.next = newNode
You have to take care of deletion manually, e.g. check if aNode.next isn't already occupied (and delete if it is).
Further, the add function signature should read:
void add(node & aNode) { ... }
By the way, the STL comes with a nice <forward_list> ;-)
It's hard to tell what you're actually asking, but going by the question title perhaps you have in mind a node structure like this:
struct Node {
Node & next;
/* payload data */
Node(Node & n) : next(n) /* ... */ { }
};
Such a node would store its successor "by reference"; but you would have to initialize it with an existing node! (There is no such thing as a "null" reference.) By the Poultry-Oval Impasse, you cannot do this.
Alright, while you continue to refuse to post your full code, here is my almost literal copy/paste of your code which works fine with me:
Update: I'm adding a feature to add a node at the end, which you might want.
#include <string>
struct node {
std::string info;
node *next;
node(std::string i = "") : info(i), next(NULL) { }
};
void add(node &aNode)
{
node *newNode;
newNode = new node;
aNode.next = newNode;
}
void add_at_end(node &aNode, std::string value = "")
{
node *newNode, *n = &aNode;
while (n->next) n = n->next; // move to the end
newNode = new node(value);
n->next = newNode;
}
int main()
{
node aNode, bNode;
add(aNode);
add_at_end(bNode, "Hello");
add_at_end(bNode, "World");
add_at_end(bNode, "!");
}
Compile with g++ -o prog prog.cpp -W -Wall -pedantic.
Finally, here's the STL way of achieving the same thing:
#include <forward_list>
#include <string>
int main() {
std::forward_list<std::string> bList;
bList.push_front("Hello");
bList.push_front("World");
bList.push_front("!");
}
In your second variant of main(), you are calling add(aNode) twice. But you're providing it the same parameter each time. So although you're creating two new node objects, one of them is lost forever (a memory leak). And aNode.next ends up pointing to the other one. aNode.next->next is not a valid pointer, hence the seg-fault when you try to access something through it.
Depending on what you want to achieve, you could try this:
node aNode;
add(aNode); // Basically does: aNode.next = new node;
add(*aNode.next); // Basically does: aNode.next->next = new node;
There are better ways of doing linked-lists, but this would at least avoid the seg-fault.
Try
int main() {
node *aNode;
aNode = new node;
add (*aNode);
}
You have to pass reference to object, not a pointer.
I checked your code and I didn't get segmentation fault when allocating on stack: http://ideone.com/gTRIG.
My proposition:
#include <string>
using namespace std;
struct node {
string info;
node *next;
node(string str): info(str), next(NULL) {}
~node() { if(next != NULL) delete next; }
node *add(string info){
node *newNode = new node(info);
return aNode.next = newNode;
}
};
int main(){
node rootNode("My rootnode");
node *nxt = rootNode.add("Next node");
nxt->add("Last node");
// No need to call delete, because destructor will clear heap
}