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

Time Complexity: Endianness (big-endian vs little-endian)

O(n)

Understanding Time Complexity

When working with endianness, we want to know how the time to convert or check byte order changes as data size grows.

We ask: How does the work grow when handling bigger numbers or more bytes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

unsigned int swap_endian(unsigned int val) {
    return ((val >> 24) & 0x000000FF) |
           ((val >> 8)  & 0x000000FF) |
           ((val << 8)  & 0x0000FF00) |
           ((val << 24) & 0xFF000000);
}
    

This code swaps the byte order of a 4-byte integer to convert between big-endian and little-endian.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Bit shifting and masking on each byte of the 4-byte integer.
• How many times: Fixed 4 times, once per byte.

How Execution Grows With Input

Each byte requires a fixed set of operations, so as the number of bytes grows, the work grows proportionally.

Input Size (bytes)	Approx. Operations
4	4 sets of shifts and masks
8	8 sets of shifts and masks
16	16 sets of shifts and masks

Pattern observation: The work grows linearly with the number of bytes processed.

Final Time Complexity

Time Complexity: O(n)

This means the time to swap bytes grows in direct proportion to the number of bytes in the data.

Common Mistake

[X] Wrong: "Swapping endianness is always a constant time operation regardless of data size."

[OK] Correct: For larger data types or arrays, each byte must be processed, so time grows with data size.

Interview Connect

Understanding how byte order conversion scales helps you write efficient embedded code and shows you can think about performance in low-level tasks.

Self-Check

"What if we needed to swap endianness for an array of integers instead of a single integer? How would the time complexity change?"

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