Memory Layout Comparison Array vs Linked List in DSA C - Build Both Approaches

Memory Layout Comparison: Array vs Linked List

📖 Scenario: Imagine you are organizing a bookshelf. You want to understand how books are stored differently when placed side-by-side on a shelf (like an array) versus when each book is kept in a separate box connected by strings (like a linked list).

🎯 Goal: You will create a simple array and a linked list with the same numbers. Then, you will print their memory addresses to see how they are stored in memory differently.

📋 What You'll Learn

Create an integer array called arr with values 10, 20, 30, 40, 50

Create a linked list with nodes containing the same values in the same order

Print the memory addresses of each element in the array

Print the memory addresses of each node in the linked list

💡 Why This Matters

🌍 Real World

Understanding memory layout helps optimize programs for speed and memory usage, important in systems programming and embedded devices.

💼 Career

Software developers and system engineers often need to choose the right data structure based on memory and performance characteristics.

1

Create an integer array with fixed values

Create an integer array called arr with exactly these values: 10, 20, 30, 40, 50.

DSA C

# Create the integer array 'arr' with the given values
// Your code here

Hint

Use the syntax int arr[5] = {10, 20, 30, 40, 50}; to create the array.

2

Define the linked list node and create the list

Define a struct called Node with an integer data and a pointer next. Then create 5 nodes with values 10, 20, 30, 40, 50 linked in order, and assign the head pointer head to the first node.

DSA C

#include 
#include 

// Define struct Node here
// Create linked list nodes and link them
// Assign head pointer

int main() {
    int arr[5] = {10, 20, 30, 40, 50};
    // Your code here
    return 0;
}

Hint

Define the struct with int data and struct Node* next. Use malloc to create nodes and link them with next.

3

Print memory addresses of array elements and linked list nodes

Use a for loop with variable i to print the memory addresses of each element in arr. Then use a while loop with a pointer current starting at head to print the memory addresses of each linked list node.

DSA C

#include 
#include 

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

int main() {
    int arr[5] = {10, 20, 30, 40, 50};

    struct Node* head = malloc(sizeof(struct Node));
    struct Node* second = malloc(sizeof(struct Node));
    struct Node* third = malloc(sizeof(struct Node));
    struct Node* fourth = malloc(sizeof(struct Node));
    struct Node* fifth = malloc(sizeof(struct Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = fourth;

    fourth->data = 40;
    fourth->next = fifth;

    fifth->data = 50;
    fifth->next = NULL;

    // Print array element addresses
    // Print linked list node addresses
    // Your code here

    return 0;
}

Hint

Use a for loop to print array addresses with &arr[i]. Use a while loop to traverse linked list nodes and print their addresses.

4

Print the final output showing memory layout comparison

Run the program to print the memory addresses of array elements and linked list nodes. The output should show contiguous addresses for the array and scattered addresses for the linked list.

DSA C

#include 
#include 

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

int main() {
    int arr[5] = {10, 20, 30, 40, 50};

    struct Node* head = malloc(sizeof(struct Node));
    struct Node* second = malloc(sizeof(struct Node));
    struct Node* third = malloc(sizeof(struct Node));
    struct Node* fourth = malloc(sizeof(struct Node));
    struct Node* fifth = malloc(sizeof(struct Node));

    head->data = 10;
    head->next = second;

    second->data = 20;
    second->next = third;

    third->data = 30;
    third->next = fourth;

    fourth->data = 40;
    fourth->next = fifth;

    fifth->data = 50;
    fifth->next = NULL;

    printf("Array element addresses:\n");
    for (int i = 0; i next;
        index++;
    }

    return 0;
}

// Run the program to see the output

Hint

Run the program to see the memory addresses printed. Notice the array addresses are close together, while linked list nodes are scattered.