Endianness (big-endian vs little-endian) in Embedded C - Hands-On Comparison

Understanding Endianness in Embedded C

📖 Scenario: You are working on a small embedded system that communicates with other devices. Different devices may use different ways to store multi-byte numbers in memory. This is called endianness.Understanding how to check and handle endianness is important to make sure data is interpreted correctly.

🎯 Goal: You will write a simple C program to detect if your system is using big-endian or little-endian format.This knowledge helps you write code that works correctly on different hardware.

📋 What You'll Learn

Create a 16-bit integer variable with a known value

Create a pointer to access the bytes of the integer

Write logic to check the first byte to determine endianness

Print the detected endianness as a string

💡 Why This Matters

🌍 Real World

Embedded systems often communicate with other devices that may use different byte orders. Detecting and handling endianness ensures data is interpreted correctly.

💼 Career

Understanding endianness is important for embedded developers, firmware engineers, and anyone working with low-level data communication or hardware interfaces.

1

DATA SETUP: Create a 16-bit integer variable

Create a 16-bit unsigned integer variable called num and set it to 0x1234.

Embedded C

# Create a 16-bit unsigned integer variable called num and set it to 0x1234
// Your code here

Hint

Use uint16_t from stdint.h to create a 16-bit unsigned integer.

2

CONFIGURATION: Create a byte pointer to the integer

Create a pointer called ptr of type uint8_t * and set it to point to num.

Embedded C

#include 
#include 

uint16_t num = 0x1234;

// Create a uint8_t pointer called ptr pointing to num
// Your code here

Hint

Cast the address of num to uint8_t * to access individual bytes.

3

CORE LOGIC: Check the first byte to detect endianness

Write an if statement that checks if the first byte pointed by ptr is 0x12. If yes, set a char pointer endian to "Big-endian", else set it to "Little-endian".

Embedded C

#include 
#include 

uint16_t num = 0x1234;
uint8_t *ptr = (uint8_t *)#

// Declare a char pointer called endian
// Use if statement to check *ptr and assign endian
// Your code here

Hint

The first byte of num will be 0x12 if big-endian, else 0x34 if little-endian.

4

OUTPUT: Print the detected endianness

Write a printf statement to print the text "System is: " followed by the value of endian.

Embedded C

#include 
#include 

uint16_t num = 0x1234;
uint8_t *ptr = (uint8_t *)#

char *endian;
if (*ptr == 0x12) {
    endian = "Big-endian";
} else {
    endian = "Little-endian";
}

// Print the detected endianness
// Your code here

Hint

Use printf("System is: %s\n", endian); inside main() to show the result.

Practice

1. What does little-endian mean in terms of byte order?

easy

A. Bytes are stored in reverse alphabetical order

B. The largest byte is stored first in memory

C. Bytes are stored in random order

D. The smallest byte is stored first in memory

Solution

1. Step 1: Understand byte order concept

   Little-endian means the smallest byte (least significant) is stored at the lowest memory address.

2. Step 2: Compare with big-endian

   Big-endian stores the largest byte first, so little-endian is the opposite.

3. Final Answer:

   The smallest byte is stored first in memory -> Option D

4. Quick Check:

   Little-endian = smallest byte first [OK]

Hint: Little-endian starts with smallest byte first [OK]

Common Mistakes:

• Confusing little-endian with big-endian
• Thinking bytes are stored randomly
• Mixing up byte significance order

2. Which of the following C code snippets correctly checks if the system is little-endian?

easy

A. int x = 1; char *c = (char*)&x; if (*c == 1) printf("Little-endian\n");

B. int x = 1; char *c = (char*)&x; if (*c == 0) printf("Little-endian\n");

C. int x = 256; char *c = (char*)&x; if (*c == 1) printf("Little-endian\n");

D. int x = 256; char *c = (char*)&x; if (*c == 0) printf("Little-endian\n");

Solution

1. Step 1: Understand the test logic

   Setting int x=1 stores 0x00000001 in memory. In little-endian, lowest byte is 1.

2. Step 2: Check pointer dereference

   Pointer to char reads first byte. If *c == 1, system is little-endian.

3. Final Answer:

   int x = 1; char *c = (char*)&x; if (*c == 1) printf("Little-endian\n"); -> Option A

4. Quick Check:

   Check first byte == 1 means little-endian [OK]

Hint: Check first byte of int 1 for 1 to detect little-endian [OK]

Common Mistakes:

• Using wrong value for x
• Checking *c == 0 instead of 1
• Confusing pointer types

3. Given the 32-bit hex number 0x12345678 stored in memory on a little-endian system, what is the byte order in memory (from lowest to highest address)?

medium

A. 12 34 56 78

B. 78 56 34 12

C. 34 12 78 56

D. 56 78 12 34

Solution

1. Step 1: Understand little-endian byte order

   Little-endian stores the least significant byte first. For 0x12345678, bytes are 12, 34, 56, 78.

2. Step 2: Arrange bytes in memory order

   Memory order is 78 (lowest address), 56, 34, 12 (highest address).

3. Final Answer:

   78 56 34 12 -> Option B

4. Quick Check:

   Little-endian reverses byte order [OK]

Hint: Little-endian reverses byte order from hex [OK]

Common Mistakes:

• Reading bytes in normal hex order
• Mixing big-endian and little-endian
• Confusing byte significance

4. The following code attempts to print the bytes of an integer in big-endian order on any system. What is the bug?

void print_big_endian_bytes(int x) {
  unsigned char *p = (unsigned char*)&x;
  for (int i = 0; i < 4; i++) {
    printf("%02x ", p[i]);
  }
  printf("\n");
}

medium

A. It prints bytes in system's native order, not big-endian

B. Pointer type is incorrect, should be int*

C. Loop runs too many times causing overflow

D. printf format specifier is wrong for bytes

Solution

1. Step 1: Analyze byte printing logic

   The code prints bytes from lowest to highest address, which is system native order.

2. Step 2: Identify missing conversion

   To print big-endian, bytes must be printed from highest to lowest address on little-endian systems.

3. Final Answer:

   It prints bytes in system's native order, not big-endian -> Option A

4. Quick Check:

   Printing p[i] from 0 to 3 prints native order [OK]

Hint: Print bytes reversed for big-endian on little-endian [OK]

Common Mistakes:

• Assuming pointer type causes bug
• Ignoring byte order differences
• Thinking printf format is wrong

5. You receive a 16-bit number from a big-endian device as two bytes: 0x01 and 0x02. Your little-endian system reads these bytes into memory in order. How do you correctly reconstruct the original number in C?

hard

A. int num = bytes[0] + bytes[1];

B. int num = (bytes[1] << 8) | bytes[0];

C. int num = (bytes[0] << 8) | bytes[1];

D. int num = bytes[1] * 256 + bytes[0];

Solution

1. Step 1: Understand byte order from big-endian device

   Big-endian sends most significant byte first: 0x01 then 0x02.

2. Step 2: Combine bytes correctly on little-endian system

   Shift first byte left 8 bits and OR with second byte: (0x01 << 8) | 0x02 = 0x0102.

3. Final Answer:

   int num = (bytes[0] << 8) | bytes[1]; -> Option C

4. Quick Check:

   Shift MSB left 8 bits, OR with LSB [OK]

Hint: Shift first byte left 8 bits, OR with second byte [OK]

Common Mistakes:

• Swapping byte order incorrectly
• Adding bytes instead of bitwise OR
• Ignoring device byte order