Endianness (big-endian vs little-endian) in Embedded C

Introduction

Endianness tells us how computers store numbers in memory. It helps us understand how data is arranged when we read or write it.

When sending data between different devices that may store numbers differently.

When reading raw data from files or network that use a specific byte order.

When debugging low-level programs that work with memory directly.

When writing code for embedded systems that communicate with hardware.

When converting data formats between systems with different endianness.

Syntax

Embedded C

/* No direct syntax in C, but we check endianness like this: */
#include <stdio.h>

int main() {
    unsigned int x = 1;
    char *c = (char*)&x;
    if (*c) {
        // little-endian
    } else {
        // big-endian
    }
    return 0;
}

Endianness is about byte order in memory, not a language keyword.

We use pointers to check how bytes are stored.

Examples

This prints the first byte of the number in memory. If it prints 0x78, the system is little-endian. If 0x12, big-endian.

Embedded C

unsigned int x = 0x12345678;
char *c = (char*)&x;
printf("First byte: 0x%x\n", (unsigned char)*c);

This simple check tells the endianness of the system.

Embedded C

unsigned int x = 1;
char *c = (char*)&x;
if (*c == 1) {
    printf("Little-endian\n");
} else {
    printf("Big-endian\n");
}

Sample Program

This program shows how the bytes of a number are stored in memory. It prints each byte and then tells if the system is little-endian or big-endian.

Embedded C

#include <stdio.h>

int main() {
    unsigned int x = 0x12345678;
    char *c = (char*)&x;

    printf("Memory bytes of 0x12345678:\n");
    for (int i = 0; i < sizeof(x); i++) {
        printf("Byte %d: 0x%x\n", i, (unsigned char)c[i]);
    }

    if (*c == 0x78) {
        printf("System is little-endian\n");
    } else if (*c == 0x12) {
        printf("System is big-endian\n");
    } else {
        printf("Unknown endianness\n");
    }

    return 0;
}

Important Notes

Most modern PCs use little-endian format.

Network protocols often use big-endian (called network byte order).

Always check endianness when sharing binary data between different systems.

Summary

Endianness is about the order of bytes in memory for numbers.

Little-endian stores the smallest byte first; big-endian stores the largest byte first.

Knowing endianness helps when working with low-level data or different devices.

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