Writing data to I2C device in Embedded C - Time & Space Complexity

Time Complexity: Writing data to I2C device

O(n)

Understanding Time Complexity

When writing data to an I2C device, it's important to understand how the time taken grows as the amount of data increases.

We want to know how the number of bytes affects the total time spent sending data.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void i2c_write_data(uint8_t device_addr, uint8_t* data, uint16_t length) {
    i2c_start();
    i2c_send_byte(device_addr << 1); // Send device address with write bit
    for (uint16_t i = 0; i < length; i++) {
        i2c_send_byte(data[i]);
    }
    i2c_stop();
}
    

This code sends a start signal, writes the device address, then sends each byte of data one by one, and finally sends a stop signal.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sending each byte of data inside the for-loop.
• How many times: The loop runs once for every byte in the data array, so length times.

How Execution Grows With Input

As the number of bytes to send grows, the total time grows in a straight line with it.

Input Size (n)	Approx. Operations
10	About 10 send operations
100	About 100 send operations
1000	About 1000 send operations

Pattern observation: Doubling the data doubles the number of send operations.

Final Time Complexity

Time Complexity: O(n)

This means the time to write data grows directly in proportion to the number of bytes you send.

Common Mistake

[X] Wrong: "Sending data to an I2C device takes the same time no matter how many bytes are sent."

[OK] Correct: Each byte requires a separate send operation, so more bytes mean more time.

Interview Connect

Understanding how data size affects communication time helps you design efficient embedded systems and answer questions about performance in real projects.

Self-Check

"What if we buffered multiple bytes before sending them all at once? How would the time complexity change?"

Practice

1. What is the first step when writing data to an I2C device in embedded C?

easy

A. Read the device address

B. Start the I2C communication

C. Stop the I2C communication

D. Send the data byte

Solution

1. Step 1: Begin communication with the device

   Writing data to an I2C device always starts with initiating the communication by sending a start condition. After starting, the device address and data can be sent, but the first step must be the start condition.

2. Final Answer:

   Start the I2C communication -> Option B

3. Quick Check:

   Start communication = first step [OK]

Hint: Always start communication before sending data [OK]

Common Mistakes:

• Sending data before starting communication
• Stopping communication too early
• Skipping the start condition

2. Which of the following is the correct way to specify the 7-bit I2C device address for writing in embedded C?

easy

A. device_address << 1 | 0

B. device_address >> 1

C. device_address << 1 | 1

D. device_address << 1 & ~1

Solution

1. Step 1: Understand I2C address format

   The 7-bit device address is shifted left by one bit to make room for the read/write bit as the least significant bit. For writing, the least significant bit must be 0, so OR with 0 is correct.

2. Final Answer:

   device_address << 1 | 0 -> Option A

3. Quick Check:

   Write bit = 0, address shifted left = correct [OK]

Hint: Shift address left and set LSB to 0 for write [OK]

Common Mistakes:

• Setting LSB to 1 for write operation
• Shifting address right instead of left
• Not shifting address at all

3. Given the following code snippet, what will be the result of the I2C write operation?

uint8_t device_addr = 0x50;
uint8_t reg_addr = 0x10;
uint8_t data = 0xAB;

I2C_Start();
I2C_Write(device_addr << 1 | 0);
I2C_Write(reg_addr);
I2C_Write(data);
I2C_Stop();

medium

A. Only register address is sent, data is missing

B. Data 0x10 is written to register 0xAB of device 0x50

C. Write operation fails due to wrong device address

D. Data 0xAB is written to register 0x10 of device 0x50

Solution

1. Step 1: Analyze the address and data sent

   The device address is correctly shifted and write bit set to 0. Then register address 0x10 and data 0xAB are sent in order. The sequence Start -> Address+Write -> Register -> Data -> Stop is correct for writing data to a register.

2. Final Answer:

   Data 0xAB is written to register 0x10 of device 0x50 -> Option D

3. Quick Check:

   Correct sequence = data written properly [OK]

Hint: Check address shift and write bit, then data order [OK]

Common Mistakes:

• Mixing register and data order
• Not shifting device address
• Forgetting to send stop condition

4. Identify the error in the following I2C write code snippet:

uint8_t device_addr = 0x68;
uint8_t reg_addr = 0x20;
uint8_t data = 0x55;

I2C_Start();
I2C_Write(device_addr >> 1 | 0);
I2C_Write(reg_addr);
I2C_Write(data);
I2C_Stop();

medium

A. Device address is shifted right instead of left

B. Data byte is sent before register address

C. Stop condition is missing

D. Write bit is set to 1 instead of 0

Solution

1. Step 1: Check device address shifting

   The device address is incorrectly shifted right instead of left, which corrupts the address sent on the bus. Register and data are sent in correct order, and stop condition is present, so no other errors.

2. Final Answer:

   Device address is shifted right instead of left -> Option A

3. Quick Check:

   Address must be shifted left for I2C write [OK]

Hint: Always shift device address left by 1 bit [OK]

Common Mistakes:

• Shifting address right instead of left
• Sending data before register address
• Omitting stop condition

5. You need to write multiple bytes to consecutive registers starting at 0x30 on an I2C device with address 0x60. Which code snippet correctly performs this operation in embedded C?

uint8_t device_addr = 0x60;
uint8_t start_reg = 0x30;
uint8_t data[] = {0x01, 0x02, 0x03};

// Which snippet is correct?

hard

A. I2C_Start();
I2C_Write(device_addr << 1 | 0);
for (int i = 0; i < 3; i++) {
I2C_Write(start_reg + i);
I2C_Write(data[i]);
}
I2C_Stop();

B. I2C_Start();
I2C_Write(device_addr << 1 | 0);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

C. I2C_Start();
I2C_Write(device_addr << 1 | 0);
I2C_Write(start_reg);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

D. I2C_Start();
I2C_Write(device_addr << 1 | 1);
I2C_Write(start_reg);
for (int i = 0; i < 3; i++) {
I2C_Write(data[i]);
}
I2C_Stop();

Solution

1. Step 1: Start communication and send device address with write bit

   The device address is shifted left and write bit set to 0, then start register address is sent once. After sending the start register, consecutive data bytes are sent to consecutive registers automatically by the device.

2. Final Answer:

   Start register sent once, then data bytes sent in loop -> Option C

3. Quick Check:

   Send start reg once, then data bytes [OK]

Hint: Send start register once, then all data bytes [OK]

Common Mistakes:

• Sending register address before each data byte
• Using read bit instead of write bit
• Not sending start register at all