pointers – Improve my solution to basic C linked list management functions

I would appreciate some help relative to my code solution, which deals with linked list management in C. I’ll already declare the only strange writing thing with my request: I am a C++ file, but I am actually mostly leveraging C resources (malloc(), free(), etc.); that said, given the basic code I provide, I am confident no one will have trouble with that.

I want to write a function to add elements to the end of the list and one to delete elements from it, that work in any edge case. Given my desire, the removal function was the one that I struggled the most with, but also the one that made me realize how little I am understanding pointers.

I will now share the code I produced, that should be working fine, but:

  • It can surely be greatly improved both in terms of clarity and performance
  • I think that showing it to the community will highlight many of the flaws present in my solution
// The plan is to create a linked list and to be able to add and delete its elements

#include <iostream>
using namespace std; // I can write output lines as cout << "Hi!", rather than std::cout < "Hi!"
#include <cstdlib> // needed for malloc() in C++

struct node {
    int data;
    node* nextPtr; //"struct node* nextPtr;" : This would be the syntax for plain old C: you always have to type the "struct" keyword
};

node* createElement(int data) {
    node* newElemPtr = (node*)malloc(sizeof(node)); // the "(node*)" cast is required by C++, and is not used in C
    newElemPtr->data = data;
    newElemPtr->nextPtr = NULL;
    return newElemPtr;
}

void appendElement(int data, node** head) { // Adds a new node at the end of the list
    // I pass as argument a pointer to pointer (double pointer) to node, so that I can edit the head node
    // if the list is empty, without having to return a new node pointer as head: my function indeed features
    // "void" in its signature
    node* elemPtr = NULL;
    elemPtr = createElement(data); // elemPtr is a pointer to the new node

    if (*head == NULL) {
        *head = elemPtr;
    }
    else {
        node* currPtr = *head; // currPtr is the temporary variable that visits each node of the linked list
        while (currPtr->nextPtr != NULL)
            currPtr = currPtr->nextPtr;
        currPtr->nextPtr = elemPtr; // Set last element's nextPtr to "elem", i.e., a pointer to the new element
    }
};

void removeElement(int data, node** head) { // Remove all the nodes whose data content matches the "data" argument
    int presence_flag = 0; // Flag used to check whether the required data is present at all in the linked list
    if (*head == NULL) {
        return;
    }
    else {
        node* currPtr = *head;
        node* prevPtr = *head;
        while (currPtr != NULL) {
            // This is the case in which I find a node to delete (it matches the "data" query), and it is not the first of the list
            if (data == currPtr->data && currPtr != *head) {
                prevPtr->nextPtr = currPtr->nextPtr; // Link the node ahead of the one to delete with the one behind
                free(currPtr);
                currPtr = prevPtr; // In the next loop, I will resume the analysis from the previous node, which now points to an unvisited one
                presence_flag = 1;
            }
            // This is the case in which I find a node to delete and it is the first of the list
            else if (data == currPtr->data && currPtr == *head) {
                // This is the case in which I have to delete the first node, but the list features other nodes
                if (currPtr->nextPtr != NULL){
                    *head = currPtr->nextPtr; // Move *head forward
                    currPtr = *head; // Do the same with currPtr, in order not to break the while() loop
                    free(prevPtr); // As *head has already been re-assigned, I leverage prevPtr to delete the old *head
                    presence_flag = 1;
                }
                // This is the case in which I have to delete the first and only node of the list
                else {
                    *head = NULL;
                    currPtr = *head;
                    presence_flag = 1;
                }
            }
            // This is the case in which the current node does not match the queried "data" value
            else{
                prevPtr = currPtr; // Update prevPtr
                currPtr = currPtr->nextPtr; // Move currPtr forward
            }
        }
    }
    if (presence_flag == 0)
        cout << "There is not any node with value " << data << " in the linked list.nn";

    // Q1: Am I causing any memory leak by using *head == NULL instead of free(*head)?
    // Q2: Should I free() everythin before ending the main(), at least as a good practice?
    // Q3: Is there a way to make this function by not using a double pointer as input and by also keeping "void" as return value?
    //     Of course, it should still work in the tricky edge case of the last element in the list that has to be deleted
};

void printLinkedList(node* head) { // Here I return nothing, so I can freely edit "head" (i.e., there is no need for a temporary pointer)
    if (head == NULL) {
        cout << "The linked list is empty.n";
    }
    else {
        int elemCounter = 0;
        while (head != NULL) {
            elemCounter += 1;
            cout << "elem N. " << elemCounter << ": data value = " << head->data << "n"; // head->data is equal to (*head).data
            head = head->nextPtr;
        }
    }
};

int main(int argc, char* argv[])
{
    //cout << "Size of a single node of the list = " << sizeof(node) << "n";
    // == 16. On a 64 bits machine, an int ("data") requires 4 bytes.
    // The pointer requires 8 bytes; the remaining 4 bytes are padding

    node* head = NULL;

    appendElement(1, &head);
    appendElement(2, &head);
    appendElement(3, &head);
    printLinkedList(head);

    cout << "nRemoving element with data value = 1...nn";
    removeElement(1, &head);
    printLinkedList(head);
    cout << "nRemoving element with data value = 2...nn";
    removeElement(2, &head);
    printLinkedList(head);
    cout << "nRemoving element with data value = 3...nn";
    removeElement(3, &head);
    printLinkedList(head);
    cout << "nRemoving element with data value = 4...nn";
    removeElement(4, &head);
    printLinkedList(head);
    cout << "nRemoving element with data value = 1...nn";
    removeElement(1, &head);
    printLinkedList(head);
    cout << "nRemoving element with data value = 2...nn";
    removeElement(2, &head);
    printLinkedList(head);

    return 0;
}

As you can see from the comments embedded in the code, I have 3 doubts that captured my interest while coding the node removal function:

  • Q1: Am I causing any memory leak by using *head == NULL instead of free(*head)?
  • Q2: Should I free() everything before ending the main(), at least as a good practice?
  • Q3: Is there a way to make this function by not using a double pointer as input and by also keeping “void” as return value? Of course, it should still work in the tricky edge case of the last element in the list that has to be deleted

I hope that featuring these “additional” questions is something reasonable to put here, as maybe someone in the future may have the same doubts I had.

I know there are plenty of ready-to-copy-and-paste solutions for my task, but I think I can really learn this stuff if I see why my precise design choices are not optimal/wrong.

I thank everyone for the time spent reading this.

Leave a Comment