Creating a Singly Linked List from Scratch in DSA C - Debugging & Practice

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this linked list traversal?

Consider a singly linked list created with nodes containing values 10, 20, and 30 in that order. What will be printed after traversing the list?

DSA C

typedef struct Node {
    int data;
    struct Node* next;
} Node;

#include <stdio.h>
#include <stdlib.h>

int main() {
    Node* head = malloc(sizeof(Node));
    head->data = 10;
    head->next = malloc(sizeof(Node));
    head->next->data = 20;
    head->next->next = malloc(sizeof(Node));
    head->next->next->data = 30;
    head->next->next->next = NULL;

    Node* current = head;
    while (current != NULL) {
        printf("%d -> ", current->data);
        current = current->next;
    }
    printf("null\n");
    return 0;
}

A 10 -> 20 -> 30 -> null

B 10 -> 20 -> null

C 30 -> 20 -> 10 -> null

D null

💡 Hint

Think about how nodes are linked and the order of traversal from head to tail.

The list is created with nodes 10, 20, and 30 linked in that order. Traversing from head prints each node's data followed by an arrow, ending with null.

🧠 Conceptual

intermediate

What happens if you forget to set the last node's next pointer to NULL?

In a singly linked list, if the last node's next pointer is not set to NULL, what is the most likely outcome when traversing the list?

A The list will automatically fix the pointer to NULL.

B The traversal will stop correctly at the last node.

C The traversal may continue into random memory causing undefined behavior or crash.

D The list will reverse itself during traversal.

💡 Hint

Think about what NULL means in a linked list.

The NULL pointer marks the end of the list. Without it, traversal continues into invalid memory, causing errors.

🔧 Debug

advanced

Why does this linked list insertion code cause a segmentation fault?

Examine the code below that tries to insert a new node at the beginning of a singly linked list. Identify the cause of the segmentation fault.

DSA C

typedef struct Node {
    int data;
    struct Node* next;
} Node;

void insertAtHead(Node* head, int value) {
    Node* newNode = malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = head;
    head = newNode;
}

int main() {
    Node* head = NULL;
    insertAtHead(head, 5);
    printf("%d\n", head->data);
    return 0;
}

A The malloc call is missing, so newNode is NULL causing a fault.

B The printf statement is incorrect and causes the fault.

C The newNode->next is not set, causing an infinite loop.

D The head pointer is passed by value, so changes inside insertAtHead do not affect the original head.

💡 Hint

Consider how pointers are passed to functions in C.

The function receives a copy of head pointer. Modifying it does not change the original head in main, so head remains NULL causing a fault when dereferenced.

📝 Syntax

advanced

Which option correctly defines a node struct for a singly linked list in C?

Choose the correct C struct definition for a singly linked list node holding an integer data.

A

struct Node {
    int data;
    Node* next;
};

B

typedef struct Node {
    int data;
    struct Node* next;
} Node;

C

typedef struct {
    int data;
    struct Node* next;
} Node;

D

typedef struct Node {
    int data;
    Node next;
} Node;

💡 Hint

Remember how to refer to the struct type inside itself.

Option B correctly uses 'struct Node* next;' inside the struct and typedefs it as Node.

🚀 Application

expert

What is the output after inserting nodes 1, 2, 3 at the head in order?

Starting with an empty singly linked list, you insert nodes with values 1, then 2, then 3 at the head. What is the final printed list?

DSA C

#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node* next;
} Node;

void insertAtHead(Node** head, int value) {
    Node* newNode = malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = *head;
    *head = newNode;
}

void printList(Node* head) {
    Node* current = head;
    while (current != NULL) {
        printf("%d -> ", current->data);
        current = current->next;
    }
    printf("null\n");
}

int main() {
    Node* head = NULL;
    insertAtHead(&head, 1);
    insertAtHead(&head, 2);
    insertAtHead(&head, 3);
    printList(head);
    return 0;
}

A 3 -> 2 -> 1 -> null

B 1 -> 2 -> 3 -> null

C null

D 1 -> null

💡 Hint

Inserting at head means new nodes become the first node.

Each new node is added at the front, so the last inserted node (3) is at the head, followed by 2, then 1.