I2C addressing (7-bit and 10-bit) in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of 7-bit I2C address shifting

What is the output of this code snippet that shifts a 7-bit I2C address to form the 8-bit address for write operation?

Embedded C

uint8_t address_7bit = 0x3A; // 7-bit address
uint8_t address_8bit = address_7bit << 1; // Shift left by 1 for write
printf("0x%02X", address_8bit);

A 0x75

B 0x3A

C 0x74

D 0x1D

💡 Hint

Remember that the 7-bit address is shifted left by 1 bit to make room for the read/write bit.

The 7-bit address 0x3A (binary 0111010) shifted left by 1 becomes 0x74 (binary 1110100). This is the 8-bit address used for write operations where the least significant bit is 0.

❓ Predict Output

intermediate

10-bit I2C address first byte calculation

Given a 10-bit I2C address 0x345, what is the value of the first byte sent on the bus according to the I2C 10-bit addressing protocol?

Embedded C

uint16_t address_10bit = 0x345;
uint8_t first_byte = 0xF0 | ((address_10bit >> 7) & 0x06); // bits 9 and 8 shifted
printf("0x%02X", first_byte);

A 0xF6

B 0xF4

C 0xF2

D 0xF0

💡 Hint

The first byte starts with 11110 and includes bits 9 and 8 of the address in bits 2 and 1.

The 10-bit address 0x345 is binary 11 0100 0101. Bits 9 and 8 are '11' which correspond to 0x06 when shifted and masked. OR with 0xF0 gives 0xF6.

🔧 Debug

advanced

Identify the error in 7-bit I2C address usage

What error will this code cause when used to send a 7-bit I2C address on the bus?

Embedded C

uint8_t address_7bit = 0x50;
uint8_t address_8bit = address_7bit | 0x01; // Attempt to set read bit
// Send address_8bit on bus

A The code will cause a compile-time error due to invalid bitwise operation

B The address is incorrect because it does not shift the 7-bit address left before adding the read bit

C The code correctly forms the 8-bit address for read operation

D The code will cause a runtime error due to overflow

💡 Hint

Think about how the 7-bit address must be shifted before adding the read/write bit.

The 7-bit address must be shifted left by 1 bit before setting the read bit. OR-ing directly with 0x01 does not shift the address, so the address bits are wrong on the bus.

📝 Syntax

advanced

Syntax error in 10-bit I2C address byte formation

Which option contains a syntax error when trying to form the second byte of a 10-bit I2C address?

Embedded C

uint16_t addr = 0x2AA;
uint8_t second_byte = addr & 0xFF;

A uint8_t second_byte = addr & 0xFF

B uint8_t second_byte = (uint8_t)(addr & 0xFF);

C uint8_t second_byte = addr & 0xFF;

D uint8_t second_byte = (addr & 0xFF);

💡 Hint

Check for missing semicolons or incorrect casting syntax.

Option A is missing the semicolon at the end of the statement, causing a syntax error.

🚀 Application

expert

Calculate total bytes sent for 10-bit I2C address write

How many bytes are sent on the I2C bus when writing to a device with a 10-bit address, including address and data bytes, if the data payload is 3 bytes?

A 3 bytes

B 4 bytes

C 6 bytes

D 5 bytes

💡 Hint

Remember that 10-bit addressing requires two address bytes plus the data bytes.

For 10-bit addressing, the master sends two address bytes (first with special bits, second with lower address bits) plus the data bytes. So 2 address bytes + 3 data bytes = 5 bytes total.

Practice

1.

What is the size of the standard I2C address used to identify devices?

easy

A. 7 bits

B. 8 bits

C. 10 bits

D. 16 bits

Solution

1. Step 1: Understand I2C addressing basics

   I2C commonly uses 7-bit addresses to identify devices on the bus.

2. Step 2: Recall standard address size

   The standard I2C address size is 7 bits, while 10-bit is an extended mode.

3. Final Answer:

   7 bits -> Option A

4. Quick Check:

   Standard I2C address = 7 bits [OK]

Hint: Remember standard I2C uses 7-bit addresses [OK]

Common Mistakes:

• Confusing 7-bit with 8-bit addresses
• Thinking 10-bit is the default
• Assuming 16-bit addresses exist in I2C

2.

Which of the following is the correct way to prepare a 7-bit I2C address for a write operation in embedded C?

uint8_t address = 0x3C; // 7-bit address
uint8_t write_address = ?;

easy

A. write_address = address << 2;

B. write_address = (address >> 1) | 1;

C. write_address = address | 0x80;

D. write_address = (address << 1) | 0;

Solution

1. Step 1: Shift 7-bit address left by 1

   The 7-bit address must be shifted left by 1 to make room for the read/write bit.

2. Step 2: Set read/write bit for write operation

   For write, the least significant bit is 0, so OR with 0.

3. Final Answer:

   write_address = (address << 1) | 0; -> Option D

4. Quick Check:

   Shift left 1 + write bit 0 = correct address [OK]

Hint: Shift left by 1, OR with 0 for write [OK]

Common Mistakes:

• Shifting right instead of left
• Setting read/write bit incorrectly
• Using wrong bitwise OR values

3.

Given the following code snippet, what is the value of header for a 10-bit I2C address 0x305?

uint16_t addr = 0x305;
uint8_t header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; // write operation

medium

A. 0xF6

B. 0xF0

C. 0xF2

D. 0xF4

Solution

1. Step 1: Analyze addr = 0x305

   Binary 0b1100000101 (bits 9-0: bit9=1, bit8=1, bit7=0). Shift right 7: 0b110 (6).

2. Step 2: Mask and compute header

   (6 & 0x06) = 0x06. Then 0xF0 | 0x06 | 0x00 = 0xF6.

3. Final Answer:

   0xF6 -> Option A

4. Quick Check:

   0xF0 | ((0x305 >> 7) & 0x06) = 0xF6 [OK]

Hint: Shift address right 7, mask with 0x06, OR with 0xF0 [OK]

Common Mistakes:

• Not shifting address before masking
• Incorrect mask value
• Confusing read/write bit in header

4.

Identify the error in this code snippet for sending a 10-bit I2C address:

uint16_t addr = 0x3FF;
uint8_t header = 0xF0 | ((addr >> 7) & 0x06) | 0x01; // read operation
uint8_t low_addr = addr & 0xFF;
// send header and low_addr over I2C

medium

A. The read/write bit in header should be 0 for read

B. The low_addr should be shifted left by 1 before sending

C. The header calculation is correct; no error

D. The mask 0x06 should be 0x07 to include all bits

Solution

1. Step 1: Check read/write bit in header

   For read operation, the last bit is set to 1, which matches the code.

2. Step 2: Verify mask and low address

   Mask 0x06 correctly extracts bits 7 and 8; low_addr is the lower 8 bits, no shift needed.

3. Final Answer:

   The header calculation is correct; no error -> Option C

4. Quick Check:

   Header and low_addr correct for 10-bit read [OK]

Hint: Check read/write bit and mask carefully [OK]

Common Mistakes:

• Confusing read/write bit value
• Incorrect mask for bits 7 and 8
• Shifting low address unnecessarily

5.

You need to write a function in embedded C that takes a 10-bit I2C address and returns the two bytes to send: the header byte and the low address byte for a write operation. Which of the following implementations is correct?

void get_i2c_10bit_bytes(uint16_t addr, uint8_t *header, uint8_t *low_addr) {
    // ?
}

hard

A. header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; low_addr = addr & 0xFF;

B. *header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; *low_addr = addr & 0xFF;

C. *header = 0xF0 | ((addr << 7) & 0x06) | 0x00; *low_addr = addr >> 8;

D. header = 0xF0 | ((addr >> 7) & 0x07) | 0x01; low_addr = addr & 0xFF;

Solution

1. Step 1: Use pointers to assign output values

   The function parameters header and low_addr are pointers, so use * to assign values.

2. Step 2: Calculate header and low_addr correctly

   Header is 0xF0 ORed with bits 7 and 8 of addr shifted right 7, and OR 0 for write; low_addr is lower 8 bits.

3. Final Answer:

   *header = 0xF0 | ((addr >> 7) & 0x06) | 0x00; *low_addr = addr & 0xFF; -> Option B

4. Quick Check:

   Pointer dereference + correct bit ops = correct function [OK]

Hint: Use * to assign pointer outputs, shift right 7 bits [OK]

Common Mistakes:

• Not using * to assign pointer values
• Shifting bits incorrectly
• Setting read/write bit wrong