Reading data from I2C device in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start an I2C transmission to the device address.

Embedded C

HAL_I2C_Master_Transmit(&hi2c1, [1], data, size, timeout);

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

A0xFF

B0x50

C0x00

D0xA0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the 7-bit address without shifting.

Using the read address instead of write address.

The device address for writing is usually the 7-bit address shifted left by 1 with the write bit (0) added. 0xA0 is a common write address for I2C devices.

2fill in blank

medium

Complete the code to read data from the I2C device.

Embedded C

HAL_I2C_Master_Receive(&hi2c1, [1], buffer, length, timeout);

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

A0xA1

B0x50

C0xA0

D0x00

Attempts:

3 left

💡 Hint

Common Mistakes

Using the write address instead of the read address.

Not shifting the address before use.

The device address for reading is the 7-bit address shifted left by 1 with the read bit (1) added. 0xA1 is a common read address.

3fill in blank

hard

Fix the error in the code to correctly read a register from the I2C device.

Embedded C

uint8_t reg = 0x10;
HAL_I2C_Master_Transmit(&hi2c1, [1], &reg, 1, timeout);
HAL_I2C_Master_Receive(&hi2c1, 0xA1, &data, 1, timeout);

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

A0x10

B0xA1

C0xA0

D0x50

Attempts:

3 left

💡 Hint

Common Mistakes

Using the read address in the transmit function.

Mixing up device addresses.

The transmit address must be the write address (0xA0) to set the register before reading.

4fill in blank

hard

Fill both blanks to read multiple bytes from a register.

Embedded C

uint8_t reg = [1];
HAL_I2C_Master_Transmit(&hi2c2, [2], &reg, 1, 100);

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

A0x20

B0xB0

C0xB1

D0x30

Attempts:

3 left

💡 Hint

Common Mistakes

Using the read address in the transmit function.

Using wrong register address.

The register to read is 0x20. The device write address is 0xB0 for hi2c2.

5fill in blank

hard

Fill all three blanks to complete the I2C read sequence for a sensor register.

Embedded C

uint8_t reg = [1];
HAL_I2C_Master_Transmit(&hi2c3, [2], &reg, 1, 100);
HAL_I2C_Master_Receive(&hi2c3, [3], &value, 1, 100);

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

A0x15

B0xC0

C0xC1

D0x16

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up read and write addresses.

Using incorrect register address.

The register to read is 0x15. The write address is 0xC0 and the read address is 0xC1 for hi2c3.

Practice

1. What is the main purpose of reading data from an I2C device in embedded C?

easy

A. To send commands to the device only

B. To get sensor or hardware information from the device

C. To reset the device hardware

D. To format the device memory

Solution

1. Step 1: Understand I2C device communication

   I2C devices often provide sensor or hardware data that the program needs to read.

2. Step 2: Identify the purpose of reading data

   Reading data means fetching information like sensor values, not sending commands or resetting.

3. Final Answer:

   To get sensor or hardware information from the device -> Option B

4. Quick Check:

   Reading I2C = Getting device data [OK]

Hint: Reading I2C means fetching data from device [OK]

Common Mistakes:

• Confusing reading with sending commands
• Thinking reading resets the device
• Assuming reading formats memory

2. Which of the following is the correct way to declare a buffer to read 4 bytes from an I2C device in embedded C?

easy

A. int buffer[1];

B. int buffer = 4;

C. char buffer;

D. char buffer[4];

Solution

1. Step 1: Understand buffer declaration for byte storage

   To store multiple bytes, we need an array of chars (bytes) of size 4.

2. Step 2: Check each option

   char buffer[4]; declares a char array of size 4, suitable for 4 bytes. Others are either single variables or wrong types.

3. Final Answer:

   char buffer[4]; -> Option D

4. Quick Check:

   Buffer for 4 bytes = char array size 4 [OK]

Hint: Use char array for byte buffers [OK]

Common Mistakes:

• Using single char instead of array
• Using int for byte buffer
• Declaring wrong array size

3. Given this code snippet, what will be the value of result after reading 2 bytes from the I2C device?

uint8_t buffer[2];
int result = i2c_read(device_addr, reg_addr, buffer, 2);

Assume i2c_read returns 0 on success and -1 on failure.

medium

A. 0

B. -1

C. 2

D. Undefined

Solution

1. Step 1: Understand i2c_read return values

   The function returns 0 if the read was successful, -1 if it failed.

2. Step 2: Analyze the code snippet

   The code reads 2 bytes and stores in buffer. If successful, result = 0.

3. Final Answer:

   0 -> Option A

4. Quick Check:

   Success return = 0 [OK]

Hint: i2c_read returns 0 on success [OK]

Common Mistakes:

• Assuming return is number of bytes read
• Confusing success with failure codes
• Ignoring return value meaning

4. Identify the error in this code snippet for reading 1 byte from an I2C device:

uint8_t data;
int status = i2c_read(device_addr, reg_addr, &data, 1);
if (status = 0) {
    process(data);
}

medium

A. Using assignment (=) instead of comparison (==) in if condition

B. Incorrect buffer type for reading data

C. Missing initialization of device_addr

D. Wrong number of bytes requested

Solution

1. Step 1: Check the if condition syntax

   The condition uses single '=' which assigns 0 instead of comparing.

2. Step 2: Understand effect of assignment in if

   Assignment returns 0 (false), so process(data) never runs even on success.

3. Final Answer:

   Using assignment (=) instead of comparison (==) in if condition -> Option A

4. Quick Check:

   Use '==' for comparison in if [OK]

Hint: Use '==' to compare, not '=' in conditions [OK]

Common Mistakes:

• Confusing '=' and '==' in if statements
• Ignoring return status check
• Not using address-of operator for buffer

5. You want to read 3 bytes from an I2C device register and combine them into a 24-bit integer. Which code snippet correctly reads and combines the data?

uint8_t buf[3];
int status = i2c_read(device_addr, reg_addr, buf, 3);
uint32_t value = ???;

hard

A. value = buf[0] | (buf[1] << 8) | (buf[2] << 16);

B. value = (buf[2] << 16) | (buf[1] << 8) | buf[0];

C. value = (buf[0] << 16) | (buf[1] << 8) | buf[2];

D. value = buf[0] + buf[1] + buf[2];

Solution

1. Step 1: Understand byte order for 24-bit integer

   Assuming buf[0] is most significant byte, shift it left by 16 bits.

2. Step 2: Combine bytes with bitwise OR

   Shift buf[1] by 8 bits, add buf[2] as least significant byte.

3. Step 3: Check each option

   value = (buf[0] << 16) | (buf[1] << 8) | buf[2]; correctly combines bytes in big-endian order. Others mix order or sum bytes incorrectly.

4. Final Answer:

   value = (buf[0] << 16) | (buf[1] << 8) | buf[2]; -> Option C

5. Quick Check:

   Shift and OR bytes for 24-bit value [OK]

Hint: Shift bytes by 16, 8, 0 bits and OR [OK]

Common Mistakes:

• Mixing byte order
• Adding bytes instead of bitwise OR
• Not shifting bytes correctly