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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a 16-bit integer variable named value.

Embedded C

uint16_t [1] = 0x1234;

Drag options to blanks, or click blank then click option'

Aval

Bvalue

Cnum

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name than 'value'.

The variable is named value as required.

2fill in blank

medium

Complete the code to create a pointer to the first byte of value.

Embedded C

uint8_t *ptr = (uint8_t *) &[1];

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Avalue

Bnum

Cdata

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name other than 'value'.

The pointer ptr points to the address of value.

3fill in blank

hard

Fix the error in the code to correctly print the first byte of value in hexadecimal.

Embedded C

printf("First byte: 0x%[1]\n", *ptr);

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AX

Bx

Cd

Do

Attempts:

3 left

💡 Hint

Common Mistakes

Using decimal (%d) or octal (%o) format specifiers.

The format specifier %X prints the number in uppercase hexadecimal.

4fill in blank

hard

Fill the blank to create a function that checks if the system is little-endian.

Embedded C

int is_little_endian() {
    unsigned int num = 1;
    return *((unsigned char *) &num) [1] 1;
}

Drag options to blanks, or click blank then click option'

A<

B!=

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using inequality or wrong comparison operators.

The function returns true if the first byte equals 1, indicating little-endian.

5fill in blank

hard

Fill all three blanks to create a function that swaps the byte order of a 16-bit integer.

Embedded C

uint16_t swap_bytes(uint16_t val) {
    return (val [1] 8) | ((val [2] 8) & [3] 0xFF00);
}

Drag options to blanks, or click blank then click option'

A<<

B>>

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong shift directions or missing masking.

The function shifts the upper byte right and the lower byte left, then masks the upper byte.

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