First post but I have been a lurker for some time now :)
First, I am attempting to execute this skeleton code in Visual Studio 2015 and it should work as is according to my teacher so I am not sure what else could be wrong client side that may be causing that.
Finally, the overall issue here is I am not sure how to complete the remain commands. I understand the basic concepts of how the pointer to pointers work as well as the linked lists but not completely. My first issue is not helping either.
Any help would be greatly appreciated.
Thanks,
Several
#include <stdio.h>
#include <stdlib.h>
typedef struct _node {
int value;
struct _node *next;
} node;
void append(node**, int);
void prepend(node**, int);
void lprint(node*);
void destroy(node*);
int main(int argc, char **argv)
{
node *head = NULL;
append(&head, 1);
append(&head, 2);
append(&head, 3);
prepend(&head, 0);
prepend(&head, -1);
prepend(&head, -2);
lprint(head);
destroy(head);
return 0;
}
void append(node **head, int value)
{
if (*head == NULL)
{
*head = (node*)calloc(0, sizeof(node));
(**head).value = value;
}
else
{
node *temp;
for (temp = *head; (*temp).next != NULL; temp = temp->next);
(*temp).next = (node*)calloc(0, sizeof(node));
(*temp).(*next).value = value;
}
}
void prepend(node **head, int value)
{
}
void lprint(node *head)
{
node *temp;
for (temp = head; temp != NULL; temp = temp->next)
{
printf("%d ", temp->value);
}
printf("\n");
}
void destroy(node *head)
{
}
I was able to compile and run your code after changing this line:
(*temp).(*next).value = value;
to this:
(*temp).next->value = value;
When I ran it, it printed out:
1 2 3
... which is what I would expect, given that prepend() isn't implemented.
I could write an implementation of prepend() and post it here, but I don't want to risk doing your homework for you; instead I'll just describe how prepend() ought to work.
For the case where head is NULL, prepend() can do the exact same thing that append() does: allocate a node and set head to point to that node.
For the other case, where head is non-NULL (because the list is non-empty), it's pretty easy too -- even easier than the append() implementation. All you need to do is allocate the new node, set the new node's next pointer to point to the existing head node (*head), and then set the head pointer (*head) to point to the new node.
The destroy() function can work with a loop very similar to the one in your lprint() function, with one caveat -- you have to grab the next pointer out of a given node (and store it into a local variable) before you free() the node, because if you free the node first and then try to read the next-pointer from the already-freed node, you are reading already-freed memory which is a no-no (undefined behavior) and will cause bad things (tm) to happen.
Related
I have been trying to understand memory allocation in C++ by reading some texts and looking stuff up. I have seen often that one should always call "delete" after "new". However, I also see code like this:
void LinkedList::add(int data){
Node* node = new Node();
node->data = data;
node->next = this->head;
this->head = node;
this->length++;
}
In structures like linked lists or stacks.
I have seen some great explanations on SO like:
Why should C++ programmers minimize use of 'new'?
When to use "new" and when not to, in C++?
However I am still confused why one would not call "delete" here for a new Node.
Edit: Let me clarify my question. I understand why not to immediately call delete in the same method. However, in the same code I do not see a matching delete statement for add. I assume that everything is deleted once the program ends, but I am confused that there is no apparent matching delete statement (ie: count all the news in the code, count all the deletes in the code, they do not match).
Edit: Here is the source that I am looking at: https://www.geeksforgeeks.org/linked-list-set-2-inserting-a-node/
The code for their linked list:
// A complete working C++ program to demonstrate
// all insertion methods on Linked List
#include <bits/stdc++.h>
using namespace std;
// A linked list node
class Node
{
public:
int data;
Node *next;
};
/* Given a reference (pointer to pointer)
to the head of a list and an int, inserts
a new node on the front of the list. */
void push(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
/* 2. put in the data */
new_node->data = new_data;
/* 3. Make next of new node as head */
new_node->next = (*head_ref);
/* 4. move the head to point to the new node */
(*head_ref) = new_node;
}
/* Given a node prev_node, insert a new node after the given
prev_node */
void insertAfter(Node* prev_node, int new_data)
{
/*1. check if the given prev_node is NULL */
if (prev_node == NULL)
{
cout<<"the given previous node cannot be NULL";
return;
}
/* 2. allocate new node */
Node* new_node = new Node();
/* 3. put in the data */
new_node->data = new_data;
/* 4. Make next of new node as next of prev_node */
new_node->next = prev_node->next;
/* 5. move the next of prev_node as new_node */
prev_node->next = new_node;
}
/* Given a reference (pointer to pointer) to the head
of a list and an int, appends a new node at the end */
void append(Node** head_ref, int new_data)
{
/* 1. allocate node */
Node* new_node = new Node();
Node *last = *head_ref; /* used in step 5*/
/* 2. put in the data */
new_node->data = new_data;
/* 3. This new node is going to be
the last node, so make next of
it as NULL*/
new_node->next = NULL;
/* 4. If the Linked List is empty,
then make the new node as head */
if (*head_ref == NULL)
{
*head_ref = new_node;
return;
}
/* 5. Else traverse till the last node */
while (last->next != NULL)
last = last->next;
/* 6. Change the next of last node */
last->next = new_node;
return;
}
// This function prints contents of
// linked list starting from head
void printList(Node *node)
{
while (node != NULL)
{
cout<<" "<<node->data;
node = node->next;
}
}
/* Driver code*/
int main()
{
/* Start with the empty list */
Node* head = NULL;
// Insert 6. So linked list becomes 6->NULL
append(&head, 6);
// Insert 7 at the beginning.
// So linked list becomes 7->6->NULL
push(&head, 7);
// Insert 1 at the beginning.
// So linked list becomes 1->7->6->NULL
push(&head, 1);
// Insert 4 at the end. So
// linked list becomes 1->7->6->4->NULL
append(&head, 4);
// Insert 8, after 7. So linked
// list becomes 1->7->8->6->4->NULL
insertAfter(head->next, 8);
cout<<"Created Linked list is: ";
printList(head);
return 0;
}
// This code is contributed by rathbhupendra
The code you cited should delete the nodes at some point. Indeed, that code shows off tons of bad C++ practices. It doesn't delete the nodes because it's bad code.
Oh and BTW: ignore anything on the site you linked to. If there is something useful on that site, it's only by accident.
Generally new does a couple of things. It allocates memory for an object, on the heap (where dynamic memory resides), and initialises an object at the address.
When you have variables in your function like this:
void example(){
int a;
char b;
}
They reside on the stack, and when the function returns, those variables no longer exist. With new you can get memory outside the stack (on the heap). The good thing is that this persists throughout function calls. The bad thing that it persists across function calls. It's good because sometimes array lengths are not known and therefore they can not be allocated on the stack, or you need a large buffer which would not fit on the stack. It's bad because if you forget about it, then it won't go away. It will just sit there taking up memory. delete, basically destructs the object at the address, and then returns the memory to the OS. That's why people say that delete should be called after new.
Luckily in modern c++ you (generally) don't need to use raw pointers and not need to worry about this. std::shared_ptr<T> created by std::make_shared<T,Args...>, and std::unique_ptr<T> created by std::make_unique<T,Args...>. These are wrappers for pointers. std::shared_ptr<T> is just T*, but when everybody loses the pointer to the object, the memory is returned. std::unique_ptr<T> is the same, but only one reference exists.
A std::unique_ptr<T> LinkedList from cppreference:
#include <memory>
struct List {
struct Node {
int data;
std::unique_ptr<Node> next;
Node(int data) : data{data}, next{nullptr} {}
};
List() : head{nullptr} {};
// N.B. iterative destructor to avoid stack overflow on long lists
~List() { while(head) head = std::move(head->next); }
// copy/move and other APIs skipped for simplicity
void push(int data) {
auto temp = std::make_unique<Node>(data);
if(head) temp->next = std::move(head);
head = std::move(temp);
}
private:
std::unique_ptr<Node> head;
};
As to an other reason the use of new should be minimised: Apart from the above issue of potential memory leak, is that it is very expensive (std::make_shared/std::make_unique still has this issue though), as the program needs to ask the kernel to grant it some memory, meaning an expensive system call must be made.
I need to write a program that takes an instance of a linear linked list and removes all of the items in the list except for the last two items. I'm doing this in c++ using classes, so I'll have 3 files: main.cpp, list.h, and list.cpp. There can be no loops: I can use multiple functions but the traversal part has to be recursive.
I thought about it and what I concluded is this: I'll have to have one public function that I can call from main which will take no arguments, called void lastTwo(). I'll have another private function called node* lastTwo(node *& current, node *& tail) which will be called by lastTwo(), and it will traverse the list recursively and delete the nodes it touches until it reaches the second to last node in the list, and then it will return the address of that node.
Then, when it returns the address of the second to last node in the list, the public function lastTwo() will catch that address and set head equal to it.
The problem I'm having is that I need to do this in vi and compile and run from the command line, and I'm getting a segmentation fault even though I drew a pointer diagram and can't find a problem with my code.
I'm working on this on the student server at my university, and all of the implementation of the data structure except for these two functions have been written by my instructors, so they're all solid. Also, every time you run the a.out file, they've written it to generate a new, random, non-empty linked list, of at least 5 nodes.
I think the problem is related to the function having the return type "node*" , because I tried doing this in visual studio as well and it wouldn't let me have functions of type node*. But, I know that when you don't use classes and just put everything in one file, functions of type node* work.
Here is my list.h:
#include<iostream>
struct node
{
int data;
node* next;
};
class list
{
public:
// these functions were already written by the school and included in the .o file
list();
~list();
void build();
void display();
// my functions
void lastTwo();
private:
node* head;
node* tail;
node* lastTwo(node *& current, node *& tail);
}
And list.cpp:
void list::lastTwo(){
node* current = head;
node * newHead = lastTwo(current, tail);
head = newHead;
delete newHead;
head->next = tail;
}
node* lastTwo(node *& current, node *& tail){
if(current->next == tail){
return current;
}
else if(current->next != tail){
node* temp = current;
current = current->next;
temp->next = NULL;
delete temp;
lastTwo(current, tail);
}
return NULL;
}
Any ideas on what might be the problem, and what the correct way to do this is, would be really appreciated! Thank you
Your problem happens when your recursion unwinds. Most of the calls to lastTwo happen in the else if. That base case is the if which returns current, but the call to lastTwo which triggered the base case is always going to return NULL when the else if ends.
Imagine this is your stack when the base case is reached:
lastTwo // Base case, returns current, but the call below this doesn't do anything with it.
lastTwo // returns null
...
lastTwo // returns null
That NULL is used by the other lastTwo and you get your seg fault.
You might use the following:
void list::lastTwo() {
// Stop work with less than 2 items
if (head == nullptr // 0 items
|| head == tail // 1 item
|| head->next == tail) { // 2 items
return;
}
// pop_front()
auto* next = head->next;
delete head;
head = next;
// loop
lastTwo();
}
I am learning list in C++ independently, and i have searched many websites about it. However, almost every approach to create a list is the same.
They usually create a struct as the node of a class. I want to create a class without using struct. So I created a class name ListNode which contains an int data and a pointer.
The main member functions of my class are AddNode and show.
Although, this program compiles successfully, it still does not work as I wish.
Here is the header file:
#ifndef LISTNODE_H_
#define LISTNODE_H_
#pragma once
class ListNode
{
private:
int data;
ListNode * next;
public:
ListNode();
ListNode(int value);
~ListNode();
void AddNode(ListNode* node,ListNode* headNode);
void show(ListNode* headNode);
};
#endif
Here is the implementation:
#include "ListNode.h"
#include<iostream>
ListNode::ListNode()
{
data = 0;
next = NULL;
}
ListNode::ListNode(int value)
{
data = value;
next = NULL;
}
ListNode::~ListNode()
{
}
void ListNode::AddNode(ListNode* node,ListNode* headNode) {
node->next = headNode;
headNode =node;
}
void ListNode::show(ListNode* headNode) {
ListNode * traversNode;
traversNode = headNode;
while (traversNode != NULL) {
std::cout << traversNode->data << std::endl;
traversNode = traversNode->next;
}
}
Main function:
#include"ListNode.h"
#include<iostream>
int main()
{
using std::cout;
using std::endl;
ListNode* head = new ListNode();
for (int i = 0;i < 3;i++) {
ListNode* Node = new ListNode(i);
head->AddNode(Node, head);
}
head->show(head);
return 0;
}
As far as I am concerned, the output should be
2
1
0
However, the output is a single zero. There must be something wrong in the AddNode and show function.
Could you please tell me what is wrong with these two functions?
When you call head->AddNode(node, head) you´re passing the memory directions which the pointers point, when the function arguments receive those directions, they are now pointing to the same directions, but those are another pointers, no the ones you declared in main. You could see it like this:
void ListNode::AddNode(ListNode* node,ListNode* headNode) {
/*when the arguments get their value it could be seen as something like:
node = Node(the one from main)
headNode = head(the one from main)*/
node->next = headNode;
/*Here you are modifying the new inserted node, no problem*/
headNode = node;
/*The problem is here, you´re modifying the memory direction
headNode points to, but the headNode argument of the function, no the one declared in main*/
}
So the pointer head in main() always points to the same first node you also declared in main().
In order to fix this you should change your code this way:
void ListNode::AddNode(ListNode* node,ListNode** headNode) {
/* second paramater now receives a pointer to apointer to a node */
node->next = *headNode;//the same as before but due to pointer syntaxis changes a bit
*headNode = node;//now you change the real head
}
And when you call it:
head->AddNode(Node, &head);//you use '&' before head
Now the real head, no the one in the function, will point to the last node you inserted.
I am writing a function to delete the last node in a linked list. This is what I have, and other code I've found online searching for a solution is very similar (I have found several), but when I execute it it creates some sort of infinite loop when deleting the last element of a linked list (it deletes other elements just fine though).
Here is the code I imagine is causing a problem:
void delete_final(Node* head){
if(head == NULL) {
return; }
if(head->next == NULL) {
delete head;
head = NULL;
return;
}
//other code
}
I imagine it's an issue with the memory (particularly after the delete head; statement), but I'm really stuck and would appreciate any help or an explanation for why this doesn't work (I possibly don't have a very good understanding of pointers and memory in C++, I'm just starting with it)
Here is my Node code for reference:
struct Node {
int key;
Node* next;
};
Thanks for any help!
Original code:
void delete_final(Node* head){
if(head == NULL) {
return; }
if(head->next == NULL) {
delete head;
head = NULL;
return;
}
//other code
}
The "other code" is not specified, but if the list has exactly one node then the above code will
delete that first node, and
update the local pointer head, which doesn't update the actual argument since it was passed by value.
As a result the calling code will be left with a dangling pointer in this case, a pointer pointing to a destroyed object, or to where such an object once was. Any use of such a pointer is Undefined Behavior. It might appear to work, or crash, or just silently cause dirty words tattoo to appear on your forehead – anything…
One fix is to pass the first-pointer by reference:
void delete_final(Node*& head){
if(head == nullptr) {
return; }
if(head->next == nullptr) {
delete head;
head = nullptr;
return;
}
//other code
}
A nice helper function for dealing with linked lists, is unlink:
auto unlink( Node*& p )
-> Node*
{
Node* const result = p;
p = p->next;
return result;
}
The implementation is perhaps a bit subtle, but all you need to remember to use it is that it updates the pointer you pass as argument, which should be either a first-node pointer or a next pointer in the list, and returns a pointer to the unlinked node.
So e.g. you can do
delete unlink( p_first );
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
class Node{
public:
int data;
Node * next;
};
void Append(Node * head, int newdata){
Node * n = head;
if(n->next != NULL)
{
n = n->next;
}
n->data = newdata;
cout<<"data-> "<<newdata<<" inserted!"<<endl;
}
void Display(Node * head){
Node * n = head;
if(n->next != NULL)
{
cout<<"data->"<<n->data<<endl;
n = n->next;
}
cout<<"data->"<<n->data<<endl;
}
int main(int argc, char * argcv[])
{
Node * headnode;
int newdata = 20;
Append(headnode, newdata);
Display(headnode);
return 0;
}
The output of this program is:
data-> 20 inserted!
data->1
data->20
Why the headnode has been assigned a "1" data member here?
Besides the comments, the problem here is that you are sending a pointer to a Node in your append function without initializing the data and the pointer that it contains.
On the if(n->next != NULL) inside append, n->next will also be random data, meaning that you will assign the value 20 not to the "head" but to the node that the head points to.
That is why the value 20 is on the next node when printing, and random value of (in this case 1) is in your head node.
You need to use the new in order to allocate the Nodes in the memory, as dereferencing a pointer that isn't pointing to allocated memory will most likely cause segfaults.
Apart from the comments and answer by #Milan, I think the design of class Node can be better implemented. It's not always required to expose data members( i.e., Node::data ) and especially in this case. class Node can also have member functions that does appending a new link and display the linked list. To get an idea, think of the way the container std::list is implemented( i.e., its member functions ). Take the advantage of powerful feature, data hiding, in C++.