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.