Memory pool (fixed-size block allocator) in Embedded C - Mini Project: Build & Apply

Memory Pool (Fixed-Size Block Allocator)

📖 Scenario: In embedded systems, memory is limited and dynamic allocation can be risky. A memory pool helps by pre-allocating fixed-size blocks of memory to manage resources efficiently and safely.Imagine you are building a small device that needs to allocate and free memory blocks quickly without fragmentation.

🎯 Goal: You will build a simple fixed-size memory pool allocator in C. It will allow allocating and freeing blocks of memory from a pre-allocated pool.This helps manage memory safely in embedded systems.

📋 What You'll Learn

Create a fixed-size memory pool array

Create a free list to track available blocks

Implement an allocate function to get a free block

Implement a free function to return a block to the pool

Print the status of the pool after allocations and frees

💡 Why This Matters

🌍 Real World

Fixed-size memory pools are used in embedded systems, real-time systems, and games to manage memory efficiently without fragmentation.

💼 Career

Understanding memory pools helps embedded developers write safer, faster code with predictable memory usage, a key skill in firmware and low-level programming jobs.

1

Create the memory pool array

Create a char array called memory_pool with size 128 to represent the fixed-size memory pool.

Embedded C

# Create a char array called memory_pool with size 128
# Your code here

Hint

Think of memory_pool as a box with 128 small slots to store data.

2

Create the free list array

Create an int array called free_list with size 16 to track free blocks. Also create an int variable called free_count and set it to 16.

Embedded C

char memory_pool[128];
# Create an int array called free_list with size 16
# Create an int variable free_count and set it to 16
# Your code here

Hint

Each block is 8 bytes (128 / 16), so free_list tracks which blocks are free.

3

Initialize the free list

Write a for loop using int i from 0 to 15 to fill free_list with block indexes from 0 to 15.

Embedded C

char memory_pool[128];
int free_list[16];
int free_count = 16;

# Use a for loop with int i from 0 to 15 to fill free_list with values 0 to 15
# Your code here

Hint

This sets all blocks as free initially.

4

Implement allocate and free functions and print status

Implement two functions: void* allocate() that returns a pointer to a free 8-byte block or NULL if none free, and void free_block(void* ptr) that returns a block to the free list. Then, allocate three blocks, free the second, and print the indexes of free blocks separated by spaces using printf.

Embedded C

char memory_pool[128];
int free_list[16];
int free_count = 16;

for (int i = 0; i < 16; i++) {
    free_list[i] = i;
}

// Implement allocate and free_block functions
// Allocate three blocks, free the second
// Print free block indexes separated by spaces
// Your code here

Hint

Remember to decrease free_count when allocating and increase it when freeing.

Printing free_list shows which blocks are free.

Practice

1.

What is the main advantage of using a memory pool with fixed-size blocks in embedded systems?

easy

A. Unlimited memory usage

B. Automatic garbage collection

C. Faster and predictable memory allocation

D. Dynamic resizing of blocks

Solution

1. Step 1: Understand memory pool purpose

   A memory pool divides memory into fixed-size blocks to allocate memory quickly without fragmentation.

2. Step 2: Identify advantage in embedded systems

   Embedded systems need fast and predictable memory allocation, which memory pools provide.

3. Final Answer:

   Faster and predictable memory allocation -> Option C

4. Quick Check:

   Memory pool advantage = Faster and predictable memory allocation [OK]

Hint: Memory pools speed up allocation by fixed block sizes [OK]

Common Mistakes:

• Confusing memory pools with dynamic resizing
• Thinking memory pools provide garbage collection
• Assuming memory pools allow unlimited memory

2.

Which of the following is the correct way to declare a fixed-size memory pool array for 10 blocks of 32 bytes each in C?

?

easy

A. char pool[10][32];

B. char pool[32][10];

C. char *pool[10];

D. char pool[10*32];

Solution

1. Step 1: Understand array dimensions

   We want 10 blocks, each block is 32 bytes, so array should be 10 rows and 32 columns.

2. Step 2: Match declaration syntax

   char pool[10][32]; declares 10 blocks of 32 bytes each correctly.

3. Final Answer:

   char pool[10][32]; -> Option A

4. Quick Check:

   Array size = 10 blocks x 32 bytes [OK]

Hint: Rows = blocks, columns = block size in bytes [OK]

Common Mistakes:

• Swapping dimensions (32 and 10)
• Using pointer array instead of fixed blocks
• Declaring a single large array without blocks

3.

Given this simplified memory pool allocator code snippet, what will be the output?

char pool[3][16];
int free_blocks = 3;

char* allocate() {
    if (free_blocks > 0) {
        return pool[--free_blocks];
    }
    return NULL;
}

int main() {
    char* p1 = allocate();
    char* p2 = allocate();
    char* p3 = allocate();
    char* p4 = allocate();
    printf("%p %p %p %p\n", p1, p2, p3, p4);
    return 0;
}

medium

A. Three valid pointers and one NULL pointer

B. All pointers are NULL

C. All pointers point to the same block

D. Compilation error due to missing headers

Solution

1. Step 1: Trace allocation calls

   Initially free_blocks=3. Each allocate() call decrements free_blocks and returns pool[free_blocks]. So p1=pool[2], p2=pool[1], p3=pool[0], p4=NULL because free_blocks=0.

2. Step 2: Understand output

   First three pointers are valid addresses, last pointer is NULL.

3. Final Answer:

   Three valid pointers and one NULL pointer -> Option A

4. Quick Check:

   3 blocks allocated, 4th returns NULL [OK]

Hint: Allocation decreases free_blocks; NULL when none left [OK]

Common Mistakes:

• Assuming pointers are all NULL
• Thinking pointers point to same block
• Forgetting to decrement free_blocks

4.

Identify the bug in this memory pool free function and select the correct fix.

void free_block(char* block) {
    if (free_blocks < MAX_BLOCKS) {
        pool[free_blocks++] = block;
    }
}

medium

A. No bug; function works correctly

B. Incrementing free_blocks before storing pointer causes overwrite

C. Should check if block is NULL before freeing

D. Store pointer in pool array is wrong; should mark block as free differently

Solution

1. Step 1: Analyze free_block logic

   The pool array holds actual memory blocks, not pointers. Storing a pointer in pool overwrites data.

2. Step 2: Correct way to free block

   Freeing should mark the block as available in a separate structure or use a free list, not overwrite pool data.

3. Final Answer:

   Store pointer in pool array is wrong; should mark block as free differently -> Option D

4. Quick Check:

   Freeing must not overwrite pool memory [OK]

Hint: Freeing must not overwrite pool data blocks [OK]

Common Mistakes:

• Overwriting pool memory with pointers
• Not checking block validity
• Assuming pool holds pointers, not blocks

5.

You want to implement a memory pool that supports 5 blocks of 64 bytes each. You also want to track which blocks are free using a bitmap (an integer where each bit represents a block's status: 0 = free, 1 = used). Which of the following code snippets correctly initializes the bitmap and allocates a block?

?

hard

A. int bitmap = 0xFF; // all used
char pool[5][64];
char* allocate() {
for (int i = 0; i < 5; i++) {
if ((bitmap & (1 << i)) == 0) {
bitmap &= ~(1 << i);
return pool[i];
}
}
return NULL;
}

B. int bitmap = 0; // all free
char pool[5][64];
char* allocate() {
for (int i = 0; i < 5; i++) {
if ((bitmap & (1 << i)) == 0) {
bitmap |= (1 << i);
return pool[i];
}
}
return NULL;
}

C. int bitmap = 0; // all used
char pool[5][64];
char* allocate() {
for (int i = 0; i < 5; i++) {
if ((bitmap & (1 << i)) != 0) {
bitmap |= (1 << i);
return pool[i];
}
}
return NULL;
}

D. int bitmap = 0xFF; // all free
char pool[5][64];
char* allocate() {
for (int i = 0; i < 5; i++) {
if ((bitmap & (1 << i)) != 0) {
bitmap &= ~(1 << i);
return pool[i];
}
}
return NULL;
}

Solution

1. Step 1: Understand bitmap meaning

   Bitmap bit 0 means free, 1 means used. So initializing bitmap=0 means all blocks free.

2. Step 2: Check allocation logic

   Allocation finds first free bit (0), sets it to 1 (used), and returns that block.

3. Final Answer:

   Bitmap initialized to 0; allocate sets bit to 1 on allocation -> Option B

4. Quick Check:

   0 = free, set bit to 1 when used [OK]

Hint: 0 bits free, set bit to 1 when allocated [OK]

Common Mistakes:

• Initializing bitmap with all bits set (all used)
• Checking wrong bit condition for free blocks
• Setting bits incorrectly on allocation