Memory allocation flow - Mini Project: Build & Apply

Memory allocation flow

📖 Scenario: Imagine you are managing a small library of books in a computer program. You want to keep track of how many books you have and store their titles. Since the number of books can change, you need to allocate memory dynamically.

🎯 Goal: You will write a C program that dynamically allocates memory to store book titles, updates the number of books, and then prints the stored titles.

📋 What You'll Learn

Create a pointer to hold dynamically allocated memory for book titles

Create a variable to store the number of books

Use malloc to allocate memory for 3 book titles

Store 3 book titles in the allocated memory

Print all stored book titles

💡 Why This Matters

🌍 Real World

Dynamic memory allocation is essential when the amount of data is not fixed, such as storing user inputs or managing collections that grow or shrink.

💼 Career

Understanding memory allocation helps in writing efficient C programs, debugging memory issues, and is crucial for systems programming and embedded development.

1

Create a pointer and a variable for books

Declare a pointer called books of type char ** to hold book titles. Also declare an integer variable called num_books and set it to 3.

C

# Declare a pointer called books and an integer num_books
# Your code here

Hint

Use char **books; to hold an array of strings and int num_books = 3; to store the count.

2

Allocate memory for book titles

Use malloc to allocate memory for num_books pointers to char and assign it to books.

C

char **books;
int num_books = 3;

# Allocate memory for books
# Your code here

Hint

Use malloc(num_books * sizeof(char *)) to allocate memory for the array of pointers.

3

Store book titles in allocated memory

Use malloc to allocate memory for each book title (assume max 20 characters). Then copy these exact titles into books[0], books[1], and books[2]: "C Programming", "Data Structures", "Algorithms".

C

char **books;
int num_books = 3;

books = (char **)malloc(num_books * sizeof(char *));

# Allocate memory for each book title and copy the titles
# Your code here

Hint

Use malloc(20 * sizeof(char)) for each title and strcpy to copy the exact strings.

4

Print all stored book titles

Use a for loop with variable i from 0 to num_books - 1 to print each book title stored in books[i].

C

char **books;
int num_books = 3;

books = (char **)malloc(num_books * sizeof(char *));

books[0] = (char *)malloc(20 * sizeof(char));
strcpy(books[0], "C Programming");

books[1] = (char *)malloc(20 * sizeof(char));
strcpy(books[1], "Data Structures");

books[2] = (char *)malloc(20 * sizeof(char));
strcpy(books[2], "Algorithms");

# Print all book titles
# Your code here

Hint

Use a for loop from 0 to num_books - 1 and printf to print each title.

Practice

1. What does the malloc function do in C programming?

easy

A. Frees previously allocated memory

B. Allocates a block of memory on the heap

C. Declares a variable

D. Copies data from one memory location to another

Solution

1. Step 1: Understand the purpose of malloc

   malloc is used to allocate memory dynamically on the heap during program execution.

2. Step 2: Differentiate from other memory functions

   free releases memory, declaration creates variables, and copying data is done by functions like memcpy.

3. Final Answer:

   Allocates a block of memory on the heap -> Option B

4. Quick Check:

   malloc = Allocate memory [OK]

Hint: Remember: malloc = get memory, free = release memory [OK]

Common Mistakes:

• Confusing malloc with free
• Thinking malloc declares variables
• Assuming malloc copies data

2. Which of the following is the correct syntax to allocate memory for 10 integers using malloc?

easy

A. int *ptr = malloc(sizeof(int));

B. int *ptr = malloc(10);

C. int ptr = malloc(10 * sizeof(int));

D. int *ptr = malloc(10 * sizeof(int));

Solution

1. Step 1: Calculate total memory needed

   To allocate memory for 10 integers, multiply 10 by the size of an int: 10 * sizeof(int).

2. Step 2: Check pointer type and assignment

   The pointer must be of type int * and assigned the result of malloc. int *ptr = malloc(10 * sizeof(int)); matches this.

3. Final Answer:

   int *ptr = malloc(10 * sizeof(int)); -> Option D

4. Quick Check:

   Correct malloc size and pointer type = int *ptr = malloc(10 * sizeof(int)); [OK]

Hint: Multiply count by sizeof(type) for malloc size [OK]

Common Mistakes:

• Not multiplying by sizeof(int)
• Using wrong pointer type
• Allocating too little memory

3. What will be the output of this C code snippet?

int *ptr = malloc(4 * sizeof(int));
if (ptr == NULL) {
    printf("Allocation failed\n");
} else {
    ptr[0] = 10;
    ptr[1] = 20;
    printf("%d %d\n", ptr[0], ptr[1]);
    free(ptr);
}

medium

A. 10 20

B. 0 0

C. Allocation failed

D. Compilation error

Solution

1. Step 1: Check memory allocation and NULL condition

   The code allocates memory for 4 integers. If allocation fails, it prints "Allocation failed". Otherwise, it proceeds.

2. Step 2: Assign and print values

   Since allocation likely succeeds, ptr[0] = 10 and ptr[1] = 20 are set, then printed as "10 20".

3. Final Answer:

   10 20 -> Option A

4. Quick Check:

   Successful malloc prints assigned values [OK]

Hint: Check if malloc returns NULL before using pointer [OK]

Common Mistakes:

• Ignoring NULL check
• Assuming uninitialized values print zero
• Forgetting to free memory

4. Identify the error in this code snippet related to memory allocation:

char *str = malloc(5);
strcpy(str, "Hello World");
free(str);

medium

A. Memory allocated is too small for the string

B. Missing check if malloc returned NULL

C. Using strcpy without including

D. No error, code is correct

Solution

1. Step 1: Check allocated memory size

   Memory allocated is 5 bytes, but the string "Hello World" needs 12 bytes (11 chars + 1 null terminator).

2. Step 2: Understand consequences of insufficient memory

   Copying a larger string into smaller allocated space causes buffer overflow and undefined behavior.

3. Final Answer:

   Memory allocated is too small for the string -> Option A

4. Quick Check:

   Allocated size < string length = Memory allocated is too small for the string [OK]

Hint: Allocate enough space including null terminator [OK]

Common Mistakes:

• Ignoring string length vs allocated size
• Not checking malloc return value
• Assuming strcpy auto-expands memory

5. You want to allocate memory for an array of 5 integers, initialize them to zero, and then safely release the memory. Which code snippet correctly implements this flow?

hard

A. int *arr = malloc(5 * sizeof(int));
for(int i=0; i<5; i++) arr[i] = 0;

B. int arr[5];
for(int i=0; i<5; i++) arr[i] = 0;

C. int *arr = calloc(5, sizeof(int));
free(arr);

D. int *arr = malloc(5);
memset(arr, 0, 5 * sizeof(int));
free(arr);

Solution

1. Step 1: Understand memory allocation and initialization

   calloc allocates and initializes memory to zero automatically, perfect for this task.

2. Step 2: Check safe memory release

   free(arr); correctly releases the allocated memory after use.

3. Final Answer:

   int *arr = calloc(5, sizeof(int)); free(arr); -> Option C

4. Quick Check:

   calloc zero-initializes and malloc does not [OK]

Hint: Use calloc for zero-initialized memory allocation [OK]

Common Mistakes:

• Using malloc without initializing memory
• Allocating too little memory
• Not freeing dynamically allocated memory