Reading data from I2C device in Embedded C - Time & Space Complexity

Time Complexity: Reading data from I2C device

O(n)

Understanding Time Complexity

When reading data from an I2C device, it's important to know how the time taken grows as we read more bytes.

We want to understand how the reading process scales with the amount of data requested.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

void read_i2c_data(uint8_t device_addr, uint8_t reg_addr, uint8_t *buffer, int length) {
    i2c_start();
    i2c_write(device_addr << 1); // Write mode
    i2c_write(reg_addr);
    i2c_start();
    i2c_write((device_addr << 1) | 1); // Read mode
    for (int i = 0; i < length; i++) {
        buffer[i] = i2c_read(i < length - 1 ? 1 : 0);
    }
    i2c_stop();
}
    

This code reads a sequence of bytes from a specific register of an I2C device into a buffer.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop reading each byte from the device.
• How many times: The loop runs exactly length times, once per byte requested.

How Execution Grows With Input

Each additional byte to read adds one more iteration of the loop, so the total time grows directly with the number of bytes.

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

Pattern observation: The time increases steadily and proportionally as more bytes are read.

Final Time Complexity

Time Complexity: O(n)

This means the time to read data grows linearly with the number of bytes requested.

Common Mistake

[X] Wrong: "Reading multiple bytes from I2C takes the same time as reading one byte."

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

Interview Connect

Understanding how data reading scales helps you design efficient embedded systems and shows you can reason about hardware communication costs.

Self-Check

"What if the I2C device supports burst reading that returns all bytes in one operation? How would the time complexity change?"

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