Fixed-width integers (uint8_t, uint16_t, uint32_t) in Embedded C - Time & Space Complexity

Time Complexity: Fixed-width integers (uint8_t, uint16_t, uint32_t)

O(n)

Understanding Time Complexity

When working with fixed-width integers like uint8_t, uint16_t, and uint32_t, it's important to understand how operations on these types affect program speed.

We want to see how the time to run code changes as the input size or number of operations grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <stdint.h>

void increment_array(uint8_t *arr, uint16_t size) {
    for (uint16_t i = 0; i < size; i++) {
        arr[i]++;
    }
}
    

This code increases each element in an array of 8-bit unsigned integers by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Incrementing each element in the array.
• How many times: Exactly once for each element, controlled by the size variable.

How Execution Grows With Input

As the array size grows, the number of increments grows at the same pace.

Input Size (n)	Approx. Operations
10	10 increments
100	100 increments
1000	1000 increments

Pattern observation: The work grows directly with the number of elements; doubling the size doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the task grows in a straight line with the number of elements.

Common Mistake

[X] Wrong: "Using smaller fixed-width integers like uint8_t makes the loop run faster because each element is smaller."

[OK] Correct: The loop runs once per element regardless of integer size; the size affects memory but not how many times the loop runs.

Interview Connect

Understanding how loops over fixed-width integers scale helps you explain performance clearly and shows you know how data size relates to processing time.

Self-Check

"What if we changed the array type from uint8_t to uint32_t but kept the same size? How would the time complexity change?"

Practice

1. What is the size in bits of the uint16_t type in embedded C?

easy

A. 64 bits

B. 8 bits

C. 32 bits

D. 16 bits

Solution

1. Step 1: Understand the meaning of uint16_t

   The suffix 16_t means it uses 16 bits to store the number.

2. Step 2: Confirm the size

   By definition, uint16_t is an unsigned integer type that uses exactly 16 bits.

3. Final Answer:

   16 bits -> Option D

4. Quick Check:

   uint16_t = 16 bits [OK]

Hint: Look at the number in the type name for bit size [OK]

Common Mistakes:

• Confusing uint16_t with uint8_t
• Thinking it stores 16 bytes instead of bits
• Mixing signed and unsigned types

2. Which of the following is the correct way to declare a 32-bit unsigned integer variable named count?

easy

A. int32_t count;

B. uint32_t count;

C. uint16_t count;

D. int count32;

Solution

1. Step 1: Identify the correct type for 32-bit unsigned integer

   uint32_t is the standard fixed-width unsigned 32-bit integer type.

2. Step 2: Check the variable declaration syntax

   The syntax uint32_t count; correctly declares the variable count with that type.

3. Final Answer:

   uint32_t count; -> Option B

4. Quick Check:

   32-bit unsigned integer declaration = uint32_t count; [OK]

Hint: Use uint32_t for 32-bit unsigned integers [OK]

Common Mistakes:

• Using signed type int32_t instead of unsigned
• Using wrong bit size like uint16_t
• Incorrect variable naming or syntax

3. What will be the output of this code snippet?

#include 
#include 

int main() {
    uint8_t a = 250;
    uint8_t b = 10;
    uint8_t c = a + b;
    printf("%u", c);
    return 0;
}

medium

A. 260

B. Overflow error

C. 4

D. 10

Solution

1. Step 1: Understand the variable types and values

   a and b are uint8_t, which can hold values 0-255. a=250, b=10.

2. Step 2: Calculate the sum and consider overflow

   Sum is 250 + 10 = 260. Since uint8_t max is 255, it wraps around modulo 256: 260 - 256 = 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   260 mod 256 = 4 [OK]

Hint: Remember 8-bit max is 255; sums wrap around modulo 256 [OK]

Common Mistakes:

• Expecting 260 as output ignoring overflow
• Thinking it causes runtime error
• Confusing signed and unsigned overflow behavior

4. Identify the error in this code snippet:

#include 

uint8_t x = 300;

medium

A. 300 is too large for uint8_t and causes overflow

B. Missing semicolon after declaration

C. uint8_t cannot be used without stdint.h

D. Variable name x is invalid

Solution

1. Step 1: Check the value assigned to uint8_t

   uint8_t can only hold values 0 to 255. 300 is outside this range.

2. Step 2: Understand the effect of assigning 300

   Assigning 300 causes overflow and the stored value wraps modulo 256, resulting in 44 (300 - 256).

3. Final Answer:

   300 is too large for uint8_t and causes overflow -> Option A

4. Quick Check:

   Value 300 > max 255 for uint8_t [OK]

Hint: Check if assigned value fits in the type's bit size [OK]

Common Mistakes:

• Assuming compiler error instead of overflow
• Ignoring the max value limit of uint8_t
• Thinking variable name or syntax is wrong

5. You want to store the number 70000 in a fixed-width unsigned integer variable. Which type should you choose to avoid overflow?

hard

A. uint32_t

B. uint16_t

C. uint8_t

D. int16_t

Solution

1. Step 1: Check the value range of each type

   uint8_t max is 255, uint16_t max is 65535, uint32_t max is 4294967295, int16_t max is 32767.

2. Step 2: Find the type that can hold 70000

   70000 is greater than 65535 and 32767, so only uint32_t can hold it without overflow.

3. Final Answer:

   uint32_t -> Option A

4. Quick Check:

   70000 <= 4294967295 of uint32_t [OK]

Hint: Pick type with max value ≥ your number to avoid overflow [OK]

Common Mistakes:

• Choosing uint16_t which is too small
• Using signed types that have smaller max positive values
• Ignoring overflow limits