Memory allocation flow

Introduction

Memory allocation flow shows how a program asks for and uses memory while running. It helps manage space for data and variables.

When you need to store data that changes size during the program.

When you want to create variables or arrays while the program runs.

When you want to use memory efficiently and free it when done.

When working with dynamic data like user input or files.

When building data structures like linked lists or trees.

Syntax

C

void* malloc(size_t size);
void free(void* ptr);

malloc asks the system for a block of memory of the given size in bytes.

free releases the memory back to the system when you no longer need it.

Examples

Allocate memory for one integer and store its address in ptr.

C

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

Allocate memory for 100 characters (like a string buffer).

C

char* buffer = (char*) malloc(100 * sizeof(char));

Release the memory previously allocated to ptr.

C

free(ptr);

Sample Program

This program allocates memory for 5 integers, stores values, prints them, and then frees the memory.

C

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

int main() {
    int* numbers = (int*) malloc(5 * sizeof(int));
    if (numbers == NULL) {
        printf("Memory allocation failed\n");
        return 1;
    }

    for (int i = 0; i < 5; i++) {
        numbers[i] = i * 10;
    }

    printf("Numbers: ");
    for (int i = 0; i < 5; i++) {
        printf("%d ", numbers[i]);
    }
    printf("\n");

    free(numbers);
    return 0;
}

Important Notes

Always check if malloc returns NULL to avoid errors.

Forgetting to free memory causes memory leaks, which waste system resources.

Memory allocated by malloc is not initialized; it may contain garbage values.

Summary

Memory allocation flow manages how programs get and release memory.

malloc allocates memory; free releases it.

Always check allocation success and free memory when done.

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