I'm attempting to craft my own basic singly linked list in C++ as a learning exercise, and I'm encountering some difficulty in the memory management department. As it stands I have...
A 'Node' class:
class Node
{
public:
char *value;
Node *next;
Node();
~Node();
};
Node::Node()
{
}
Node::~Node()
{
delete[] value;
}
And then my list (I've omitted certain method calls for brevity):
class LinkedList
{
private:
Node *head;
public:
LinkedList();
~LinkedList();
void Add(char **x);
};
LinkedList::LinkedList()
{
head = 0;
}
LinkedList::~LinkedList()
{
Node *temp;
Node *current = head;
while(current)
{
temp = current;
current = current->next;
delete temp;
}
}
void LinkedList::Add(char **x)
{
Node *nodeToAdd = new Node();
nodeToAdd->value = *x;
nodeToAdd->next = NULL;
Node *current = head;
if(!head)
{
head = nodeToAdd;
return;
}
while(current->next)
{
current = current->next;
}
current->next = nodeToAdd;
}
I'm attempting to use this code as follows (again I've omitted things for brevity):
int main()
{
LinkedList *list = new LinkedList();
char *alpha = "alpha";
char *beta = "beta";
char *charlie = "charlie";
char *delta = "delta";
char *echo = "echo";
list->Add(&alpha);
list->Add(&beta);
list->Add(&charlie);
list->Add(&delta);
list->Add(&echo);
delete list;
}
The last call in main to delete the list produces an error:
Debug Assertion Failed! Expression: _BLOCK_TYPE_IS_VALID(pHead->nBlockUse)
What am I doing wrong here?
The data pointed to by the various Node::value aren't dynamically allocated, so you shouldn't delete them. Applying the concept of "ownership", nodes should either make their own copies of data, which they own and can delete, or nodes don't own data, so they shouldn't be responsible for deleting it.
You can also implement multiple ownership using reference counting, like Objective-C does (see Objective-C Memory Management Rules for more info) but you have to be careful to avoid ownership cycles. You often find some type of reference counting in third-party smart pointers, such as Boost's smart_ptr library. Since you're doing this for the learning experience, it may make more sense to roll your own than use a library. Of course, you could also use a library for now, letting you focus on whatever you're trying to learn.
One day a student came to Moon and said: “I understand how to make a better garbage collector. We must keep a reference count of the pointers to each cons.”
Moon patiently told the student the following story:
“One day a student came to Moon and said: ‘I understand how to make a better garbage collector...
you are trying to release the memory which is not allocated on heap.
char *alpha = "alpha"; --- not allocated on heap
calling delete[]in Node destructor would lead to heap corruption.
Some points:
1) initialize pointers properly in the constructor:
Node::Node():value(NULL),next(NULL)
{
}
2) Take a ownership of value.
Allocate the memory on heap and copy
the contents
You shouldn't release a pointer use delete[]/delete if it's not created by new operator. There are some actions under the hood for the delete[] operation, like releasing/reclaiming marked memory from a managed pool. Since your pointer doesn't belong to these stuff, there will be a problem. IMHO, the underlying delete[] code is the _BLOCK_TYPE_IS_VALID(pHead->nBlockUse) stuff.
The problem is that you're assuming that you can delete the data inside node, but you're passing in pointers to string literals instead, which you can't delete.
If you're assuming that the Node object controls the lifetime of the data inside it, your Node constructor or the Add function in LinkedList will have to make a copy of the data that it is being passed.
In your destructor, you are trying to array delete (delete [ ]) a static string. You have change your Add function to reserve the string and copy it first. See the code below.
However, if I were you and fairly new to memory management, I'd really use something like CString instead of a raw "char *" as it's much easier to deal with.
void LinkedList::Add(const char *x)
{
Node *nodeToAdd = new Node();
int len=strlen(x);
nodeToAdd->value = new char [len+1]; // room for string + terminating 0
strcpy(nodeToAdd->value,x);
nodeToAdd->next = NULL;
Node *current = head;
if(!head)
{
head = nodeToAdd;
return;
}
while(current->next)
{
current = current->next;
}
current->next = nodeToAdd;
}
value and next in Node class doesn't have memory allocated. You should allocate memory in Node's constructor.
Related
I've started to learn linked lists today, and I am trying to delete nodes.
void deleteEnd(Node* refNode) {
Node* lastNode;
lastNode = new Node;
while((refNode->next)->next != NULL) {
refNode = refNode->next;
}
lastNode = refNode->next;
refNode->next = NULL;
delete lastNode;
}
void deleteIndex(Node* refNode, int index) {
Node *prev, *next, *deleted;
prev = new Node;
next = new Node;
deleted = new Node;
for(int i=1; i < index; i++) {
refNode = refNode->next;
}
prev = refNode;
deleted = prev->next;
next = deleted->next;
prev->next = next;
free(deleted);
}
I can use delete in the first one, but when I try to use it in the second, it doesn't work. The terminal doesn't give any error messages.
I found some information on the Internet, but I couldn't really understand it.
This is my linked list:
class Node {
public:
int data;
Node *next;
};
As pointed out by the comments, there are several things wrong with this code. All issues are from the comments, none are found by me, all credit goes to François Andrieux, Jesper Juhl, Sven Nilsonn, Avi Berger, and Thomas Matthews.
First, the code probably doesn't work because you mixed new and free. new is a C++ API function, while free is from C. Whenever you construct an object with new, which should not be that often with C++'s automatic memory management, you must free it with delete.
Second, when looping through a list, always start at 0. The only reason otherwise would be to start at the second item.
Third, in this passage:
prev = new Node;
...
prev = refNode;
...
prev->next = next;
When you set prev, it is overwriting the previous value. If this is a pointer, as it is, then this causes a memory leak. Always delete it before overwriting.
Finally, in deleteEnd, as pointed out by Thomas Matthews, you are trying to dereference, or get the value from, the pointers, without checking if it is nullptr. If one is, it will cause undefined behavior, and can crash the program.
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.
I have a Node list that each Node contains a pointer pointing to a Student variable (this is a class), and a pointer pointing to next Node. Here is my code for insertAtTail.
void studentRoll::insertAtTail(const Student &s) {
if (head == NULL) {
this->head = new Node;
this->head->next = NULL;
this->head->s = new Student(s);
this->tail = head;
}
else {
this->tail->next = new Node;
this->tail = this->tail->next;
this->tail->next = NULL;
this->tail->s = new Student(s);
}
}
I used valgrind to debug, and I got:
==11106== 16 bytes in 1 blocks are definitely lost in loss record 1 of 2
==11106== at 0x4C2D1CA: operator new(unsigned long)
(vg_replace_malloc.c:334)
==11106== by 0x402BE7: StudentRoll::insertAtTail(Student const&)
(studentRoll.cpp:15)
==11106== by 0x401CF1: main (testStudentRoll01.cpp:19)
==11106==
==11106== 16 bytes in 1 blocks are definitely lost in loss record 2 of 2
==11106== at 0x4C2D1CA: operator new(unsigned long)
(vg_replace_malloc.c:334)
==11106== by 0x402C5B: StudentRoll::insertAtTail(Student const&)
(studentRoll.cpp:22)
==11106== by 0x401E2C: main (testStudentRoll01.cpp:27)
==11106==
Can someone help me with it? I think there are some problems about:
this->head->s = new Student(s);
and
this->tail->s = new Student(s);
But I cannot delete them because I need these "Students."
And there are pointers point to "Students."
Thanks!!
Update: here is my destructor
StudentRoll::~StudentRoll() {
Node *iter = head;
while (iter) {
Node *next = iter->next;
iter->s->~Student();
delete iter;
iter = next;
}
head = tail = NULL;
}
Can someone help me with it? I think there are some problems about:
this->head->s = new Student(s);
and
this->tail->s = new Student(s);
But I cannot delete them because I need these "Students." And there are pointers point to "Students."
This problem probably indicates that you should redesign your program. In C++, you should express ownership semantics and make clear what objects own what resources and are responsible for their cleanup. Ownership semantics in C++ are expressed via various pointer types:
If a certain, single object owns some heap memory, rather than using raw pointers and new and delete directly, use std::unique_ptr. std::unique_ptr is better because it conveys your intent to readers and uses RAII to help prevent memory leaks.
On the other hand, if an object does not own a piece of memory, use a reference or a raw pointer instead. (In the future, the C++ standard library may get a non-owning smart pointer.)
If your linked list data structure owns the student objects, it should be the one to deallocate them. In this case, use std::unique_ptr:
void studentRoll::insertAtTail(const Student &s) {
if (head.get() == nullptr) {
this->head = std::make_unique<Node>();
this->head->next = nullptr;
this->head->s = std::make_unique<Student>(s);
this->tail = &*head; // Get a raw pointer
}
else {
this->tail->next = std::make_unique<Node>();
this->tail = &*this->tail->next; // Get a raw pointer
this->tail->next = nullptr;
this->tail->s = std::make_unique<Student>(s);
}
}
Instead of using std::unique_ptr, another option would be to simply make Student a data member of your Node type. However, this decision may indicate different intent and have different implications. For example, if you wanted to transfer ownership of the Student object from the Node object to somewhere else, you should use std::unique_ptr. If you kept the Student object directly as a member, you might achieve a similar effect by calling the Student's move constructor, but some of the semantics would still be different. For example, pointers to the Student would be invalidated. See https://stackoverflow.com/a/31724938/8887578 for more comparisons of the two approaches.
If the student objects outlive the linked list, it should not be their owner and would be better off taking a non-owning pointer to such an object. In this case, do not allocate new student objects, but take a pointer from somewhere else:
void studentRoll::insertAtTail(const Student* s) {
if (head.get() == nullptr) {
this->head = std::make_unique<Node>();
this->head->next = nullptr;
this->head->s = s;
this->tail = &*head;
}
else {
this->tail->next = std::make_unique<Node>();
this->tail = &*this->tail->next;
this->tail->next = nullptr;
this->tail->s = s;
}
}
I don't know the context of your program (e.g. if it is a school exercise for writing linked lists), but in serious code, you should use the standard library's std::list instead of rolling your own linked list. In many cases, however, a std::vector (similar to dynamically-grown array) is more appropriate than a linked list.
Additionally, instead of giving Node a parameterless, default constructor and then later assigning its s member, you should pass the student pointer to it in its constructor.
I'm creating something similar to structure list. At the beginning of main I declare a null pointer. Then I call insert() function a couple of times, passing reference to that pointer, to add new elements.
However, something seems to be wrong. I can't display the list's element, std::cout just breaks the program, even though it compiler without a warning.
#include <iostream>
struct node {
node *p, *left, *right;
int key;
};
void insert(node *&root, const int key)
{
node newElement = {};
newElement.key = key;
node *y = NULL;
std::cout << root->key; // this line
while(root)
{
if(key == root->key) exit(EXIT_FAILURE);
y = root;
root = (key < root->key) ? root->left : root->right;
}
newElement.p = y;
if(!y) root = &newElement;
else if(key < y->key) y->left = &newElement;
else y->right = &newElement;
}
int main()
{
node *root = NULL;
insert(root, 5);
std::cout << root->key; // works perfectly if I delete cout in insert()
insert(root, 2);
std::cout << root->key; // program breaks before this line
return 0;
}
As you can see, I create new structure element in insert function and save it inside the root pointer. In the first call, while loop isn't even initiated so it works, and I'm able to display root's element in the main function.
But in the second call, while loop already works, and I get the problem I described.
There's something wrong with root->key syntax because it doesn't work even if I place this in the first call.
What's wrong, and what's the reason?
Also, I've always seen inserting new list's elements through pointers like this:
node newElement = new node();
newElement->key = 5;
root->next = newElement;
Is this code equal to:
node newElement = {};
newElement.key = 5;
root->next = &newElement;
? It would be a bit cleaner, and there wouldn't be need to delete memory.
The problem is because you are passing a pointer to a local variable out of a function. Dereferencing such pointers is undefined behavior. You should allocate newElement with new.
This code
node newElement = {};
creates a local variable newElement. Once the function is over, the scope of newElement ends, and its memory gets destroyed. However, you are passing the pointer to that destroyed memory to outside the function. All references to that memory become invalid as soon as the function exits.
This code, on the other hand
node *newElement = new node(); // Don't forget the asterisk
allocates an object on free store. Such objects remain available until you delete them explicitly. That's why you can use them after the function creating them has exited. Of course since newElement is a pointer, you need to use -> to access its members.
The key thing you need to learn here is the difference between stack allocated objects and heap allocated objects. In your insert function your node newElement = {} is stack allocated, which means that its life time is determined by the enclosing scope. In this case that means that when the function exits your object is destroyed. That's not what you want. You want the root of your tree to stored in your node *root pointer. To do that you need to allocate memory from the heap. In C++ that is normally done with the new operator. That allows you to pass the pointer from one function to another without having its life time determined by the scope that it's in. This also means you need to be careful about managing the life time of heap allocated objects.
Well you have got one problem with your Also comment. The second may be cleaner but it is wrong. You have to new memory and delete it. Otherwise you end up with pointers to objects which no longer exist. That's exactly the problem that new solves.
Another problem
void insert(node *&root, const int key)
{
node newElement = {};
newElement.key = key;
node *y = NULL;
std::cout << root->key; // this line
On the first insert root is still NULL, so this code will crash the program.
It's already been explained that you would have to allocate objects dynamically (with new), however doing so is fraught with perils (memory leaks).
There are two (simple) solutions:
Have an ownership scheme.
Use an arena to put your nodes, and keep references to them.
1 Ownership scheme
In C and C++, there are two forms of obtaining memory where to store an object: automatic storage and dynamic storage. Automatic is what you use when you declare a variable within your function, for example, however such objects only live for the duration of the function (and thus you have issues when using them afterward because the memory is probably overwritten by something else). Therefore you often must use dynamic memory allocation.
The issue with dynamic memory allocation is that you have to explicitly give it back to the system, lest it leaks. In C this is pretty difficult and requires rigor. In C++ though it's made easier by the use of smart pointers. So let's use those!
struct Node {
Node(Node* p, int k): parent(p), key(k) {}
Node* parent;
std::unique_ptr<Node> left, right;
int key;
};
// Note: I added a *constructor* to the type to initialize `parent` and `key`
// without proper initialization they would have some garbage value.
Note the different declaration of parent and left ? A parent owns its children (unique_ptr) whereas a child just refers to its parent.
void insert(std::unique_ptr<Node>& root, const int key)
{
if (root.get() == nullptr) {
root.reset(new Node{nullptr, key});
return;
}
Node* parent = root.get();
Node* y = nullptr;
while(parent)
{
if(key == parent->key) exit(EXIT_FAILURE);
y = parent;
parent = (key < parent->key) ? parent->left.get() : parent->right.get();
}
if (key < y->key) { y->left.reset(new Node{y, key}); }
else { y->right.reset(new Node{y, key}); }
}
In case you don't know what unique_ptr is, the get() it just contains an object allocated with new and the get() method returns a pointer to that object. You can also reset its content (in which case it properly disposes of the object it already contained, if any).
I would note I am not too sure about your algorithm, but hey, it's yours :)
2 Arena
If this dealing with memory got your head all mushy, that's pretty normal at first, and that's why sometimes arenas might be easier to use. The idea of using an arena is pretty general; instead of bothering with memory ownership on a piece by piece basis you use "something" to hold onto the memory and then only manipulate references (or pointers) to the pieces. You just have to keep in mind that those references/pointers are only ever alive as long as the arena is.
struct Node {
Node(): parent(nullptr), left(nullptr), right(nullptr), key(0) {}
Node* parent;
Node* left;
Node* right;
int key;
};
void insert(std::list<Node>& arena, Node *&root, const int key)
{
arena.push_back(Node{}); // add a new node
Node& newElement = arena.back(); // get a reference to it.
newElement.key = key;
Node *y = NULL;
while(root)
{
if(key == root->key) exit(EXIT_FAILURE);
y = root;
root = (key < root->key) ? root->left : root->right;
}
newElement.p = y;
if(!y) root = &newElement;
else if(key < y->key) y->left = &newElement;
else y->right = &newElement;
}
Just remember two things:
as soon as your arena dies, all your references/pointers are pointing into the ether, and bad things happen should you try to use them
if you ever only push things into the arena, it'll grow until it consumes all available memory and your program crashes; at some point you need cleanup!
I'm trying to learn C++ and there is a small confusion I have.
The text which I am learning from tells me that if I want to delete a node of type const T& I should first create a new pointer of that node type, then delete it using the inbuilt C++ delete[]. However, what happens if I just set the link from the to-be-deleted node's previous element to the to-be-deleted node's next element? Something like:
*p = node.previous;
p-> next = node.next;
Or will this cause a memory leak?
I'm confused because I read somewhere else to never, ever delete pointers willy-nilly, but the example code I am working with has something along the lines of:
Node<T> *p = node-to-be-deleted;
delete p;
What is the best way to delete the node?
Assuming your node looks like this:
struct Node
{
Node* previous;
Node* next;
SomeType data;
};
Then:
*p = node.previous;
p-> next = node.next;
Then YES. This will cause a memory leak.
It also leaves p->next->prev pointing at the wrong node.
I'm confused because I read somewhere else to never, ever delete pointers willy-nilly, but the example code I am working with has something along the lines of:
Yes the best way is to "never delete pointers". But this has to go along with some context. You should not be deleting pointers manually because pointers should be managed by an objects that control their lifespan. The simplest of these objects are smart pointers or containers. But for this situation that would be overkill (as you are creating the container).
As you are creating the container (a list) you will need to do the management yourself (Note C++ already has a couple of lost types std::list for a list of values of type t or boost::ptr_list for a list of pointers to T). But it is a good exercise to try and do it yourself.
Here is an example on code review of a beginner making a list and the comments it generated:
http://codereview.stackexchange.com: Linked list in C++
I hope this helps in explains on how to create and delete objects.
Node* p = new Node; // This is how you allocate a node
delete p; // This is how you delete it
The delete[] operator should be used on dynamically allocated arrays:
Node* nodelist = new Node[ 4 ]; // nodelist is now a (dynamically allocated) array with 4 items.
delete[] nodelist; // Will delete all 4 elements (which is actually just one chunk of memory)
Deleting a Node directly only makes sense if Node implements a destructor to update the previous and next pointers of the surrounding Node instances, eg:
Node::~Node()
{
if (previous) previous->next = next;
if (next) next->previous = previous;
}
Node *p = node-to-be-deleted;
delete p;
Otherwise, you have to update the Node pointers before then deleting the Node in question, eg:
Node *p = node-to-be-deleted;
if (p->previous) p->previous->next = p->next;
if (p->next) p->next->previous = p->previous;
delete p;
With that said, the best approach is to no implement a linked list manually to begin with. In C++, use a std::list container instead, and let it handle these details for you.
void deleteNode( Node * p )
{
Node * temp = p->next;
p->data = p->next->data;
p->next = temp->next;
free(temp);
}
Heres something i did a few months ago.
template <class T>
T LinkedList<T>::remove(int pos)
{
if (pos < 1 || pos > size)
{
throw pos;
}
ListNode * temp;
if (pos == 1)
{
temp=head;
head = head->next;
}
else
{
int i=1;
ListNode * prev = head;
while(i<pos-1)
{
i++;
prev=prev->next;
}
temp = prev->next;
prev->next = (prev->next)->next;
}
--size;
return temp->item;
}