singly linked list c++ constructor, destructor and printing out

singly linked list c++ constructor, destructor and printing out

By : Ali Ramzan
Date : November 25 2020, 04:01 AM
I hope this helps . I am a beginner learning c++, and currently making a singly linked list. I have faced some problems and I thought for a very long time, searched a lot but still do not have an answer for this code so I am begging for some help.. , You could cover the issue with a pointer to pointer:
code :
List(T* arr, int n_nodes)
    Node<T>** tmp = &head; // tmp *pointing* to uninitialized(!) head pointer
    for(int i = 0; i < n_nodes; i++)
        Node<T>* node = new Node<T>();
        node->data = arr[i];
        // now the trick:
        *tmp = node; // !!!
        // you now have assigned the new node to whatever pointer
        // the tmp pointer points to - which initially is - guess - head...

        // but we now need to advance!
        tmp = &node->next;
    // tmp now points to latestly created node's next pointer
    // (or still head, if no nodes where created because of n_nodes == 0)
    // be aware that this one still is not initialized! so:
    *tmp = nullptr;
Node<T>* ptr, tmp;
for(ptr = head->next; ptr == NULL; ptr = head->next)
    delete ptr; // you delete ptr, but advancing (ptr = head->next)
                // is done AFTERWARDS, so you'd access already deleted memory
                // undefined behaviour
    Node<T>* tmp = head; // need a copy of pointer
    head = head->next;   // need to advance BEFORE deleting
    delete tmp;          // now can delete safely

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Singly linked list C, printing

Singly linked list C, printing

By : Witek
Date : March 29 2020, 07:55 AM
Any of those help You should format your code properly. first->data is allocated via malloc() and isn't initialized, so using its value invokes undefined behavior. In order not to deal the first element specially, you should use pointer to pointer to have createlist() modify first. Since createlist() won't return anything, type of its return value should be void. I guess you wanted to copy the strings instead of assigning the first character of each strings. To print all of what you entered, code to do so have to be written. You shouldn't use gets(), which has unavoidable risk of buffer overrun. You should free() whatever you allocated via malloc().
improved code:
code :
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node
    char *data;
    struct Node *next;
} node;

void createlist(node **pointer, char data[100])
    while (*pointer != NULL)
        pointer = &(*pointer)->next;

    *pointer = malloc(sizeof(node));
    if (*pointer == NULL)
        perror("malloc 1");
    (*pointer)->data = malloc(strlen(data) + 1);
    if ((*pointer)->data == NULL)
        perror("malloc 2");
    strcpy((*pointer)->data, data);
    (*pointer)->next = NULL;

int main(void)
    node *first = NULL;

    printf("Enter the lines\n");
    while (1)
        char data[100], *lf;
        if (fgets(data, sizeof(data), stdin) == NULL) strcpy(data, "quit");
        if ((lf = strchr(data, '\n')) != NULL) *lf = '\0'; /* remove newline character */
        createlist(&first, data);
        if (strcmp(data, "print") == 0)
            node *elem = first;
            while (elem != NULL)
                printf("%s\n", elem -> data);
                elem = elem->next;
        else if (strcmp(data, "quit") == 0)
            while (first != NULL)
                node *next = first->next;
                first = next;


Singly Linked List - Segmentation Error due to Destructor implementation

Singly Linked List - Segmentation Error due to Destructor implementation

By : Snir Morlevi
Date : March 29 2020, 07:55 AM
I wish this help you I'm trying to figure out why I'm getting a seg-error from my singly linked list implementation. , You have UB right here:
code :
std::unique_ptr<Node> new_node;
new_node->val = move(a);
std::unique_ptr<Node> new_node = std::make_unique<Node>(); // C++14 or later
std::unique_ptr<Node> new_node( new Node ); // pre C++14
delete &old;
std::unique_ptr<Node> tail ;
What is the proper way to delete a singly-linked-list in destructor?

What is the proper way to delete a singly-linked-list in destructor?

By : keith
Date : March 29 2020, 07:55 AM
may help you .
As you can see if I just use delete next; in destructor then I get it called as many as nodes in the list
Printing a Singly Linked List

Printing a Singly Linked List

By : Jill Rivas
Date : March 29 2020, 07:55 AM
like below fixes the issue What you need is a stack. Your system provides one automatically for you when you call a function recursively:
How is this singly linked list destructor causing an infinite loop?

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

By : kmcoder
Date : March 29 2020, 07:55 AM
Hope this helps 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.
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