Signed vs unsigned behavior on 8-bit MCU in Embedded C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an 8-bit unsigned variable.

Embedded C

unsigned char [1] = 255;

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

Anum

Bvalue

Ccount

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using a signed type instead of unsigned.

Choosing a variable name that is not descriptive.

We declare an 8-bit unsigned variable named num to hold values from 0 to 255.

2fill in blank

medium

Complete the code to assign a negative value to a signed 8-bit variable.

Embedded C

signed char [1] = -128;

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

Adata

Btemp

Cval

Dresult

Attempts:

3 left

💡 Hint

Common Mistakes

Using unsigned char for negative values.

Assigning values outside the range -128 to 127.

The variable temp is declared as signed char to hold negative values down to -128.

3fill in blank

hard

Fix the error in the code to correctly add two unsigned 8-bit variables without overflow.

Embedded C

unsigned char a = 200;
unsigned char b = 100;
unsigned int sum = a [1] b;

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

A-

B*

C+

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Storing the result in an 8-bit variable causing overflow.

Use the + operator to add the two unsigned variables. The sum is stored in a wider type to avoid overflow.

4fill in blank

hard

Fill in the blank to correctly check if a signed 8-bit variable is negative.

Embedded C

signed char x = -5;
if (x [1] 0) {
    // x is negative
}

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

A!=

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < for negative check.

Using wrong operators like == or !=.

Use < to check if x is less than zero.

5fill in blank

hard

Fill all three blanks to create a loop that counts down an unsigned 8-bit variable from 10 to 0.

Embedded C

unsigned char i = [1];
while (i [2] 0) {
    // do something
    i [3] 1;
}

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

A10

B>

C-=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > in the loop condition.

Using ++ instead of -= to decrement.

Initialize i to 10, loop while i > 0, and decrement i by 1 each time using -=.

Practice

1. What is the range of values for an 8-bit signed char on a typical MCU?

easy

A. -128 to 127

B. 0 to 255

C. -256 to 255

D. -127 to 128

Solution

1. Step 1: Understand 8-bit signed char range

   An 8-bit signed char uses one bit for sign and 7 bits for value, so it can represent values from -128 to 127.

2. Step 2: Compare with options

   Only -128 to 127 matches the correct range for signed 8-bit char.

3. Final Answer:

   -128 to 127 -> Option A

4. Quick Check:

   Signed 8-bit char range = -128 to 127 [OK]

Hint: Signed 8-bit char always includes negative half [OK]

Common Mistakes:

• Confusing signed and unsigned ranges
• Assuming 8-bit signed char goes to 255
• Mixing up sign bit usage

2. Which of the following declarations correctly defines an unsigned 8-bit variable in embedded C?

easy

A. signed char var;

B. int8_t var;

C. char unsigned var;

D. unsigned char var;

Solution

1. Step 1: Identify unsigned 8-bit type syntax

   In embedded C, unsigned char is the standard way to declare an unsigned 8-bit variable.

2. Step 2: Check other options

   signed char var; is signed, char unsigned var; is invalid syntax, int8_t var; is signed 8-bit integer.

3. Final Answer:

   unsigned char var; -> Option D

4. Quick Check:

   Unsigned 8-bit variable = unsigned char [OK]

Hint: Use 'unsigned char' for unsigned 8-bit variables [OK]

Common Mistakes:

• Using 'char unsigned' which is invalid syntax
• Assuming 'int8_t' is unsigned
• Confusing signed and unsigned keywords

3. Consider this code snippet on an 8-bit MCU:

unsigned char a = 250;
signed char b = (signed char)a;
printf("%d", b);

What is the output?

medium

A. 250

B. -6

C. 6

D. -250

Solution

1. Step 1: Understand casting from unsigned to signed

   Variable a holds 250 (unsigned). Casting to signed char wraps around because signed char max is 127.

2. Step 2: Calculate signed char value

   250 - 256 = -6 (wrap-around in 8-bit signed range)

3. Final Answer:

   -6 -> Option B

4. Quick Check:

   Unsigned 250 cast to signed 8-bit = -6 [OK]

Hint: Subtract 256 if unsigned > 127 when casting to signed [OK]

Common Mistakes:

• Assuming cast keeps original unsigned value
• Confusing sign extension with wrap-around
• Ignoring 8-bit overflow behavior

4. This code runs on an 8-bit MCU:

unsigned char x = 200;
signed char y = -100;
unsigned char z = x + y;
printf("%u", z);

What is the output?

medium

A. 156

B. 44

C. 100

D. 300

Solution

1. Step 1: Understand integer promotion in addition

   In expressions, unsigned char and signed char are promoted to int before addition.

2. Step 2: Calculate the sum

   200 + (-100) = 100 as int, but when assigned to unsigned char (8-bit), it wraps modulo 256. Since 100 fits in 8 bits, the value is 100.

3. Step 3: Print the result

   printf with %u prints unsigned int, so output is 100.

4. Final Answer:

   100 -> Option C

5. Quick Check:

   Signed promoted to int before addition, result 100 [OK]

Hint: Signed promoted to int before addition, watch final cast [OK]

Common Mistakes:

• Adding raw unsigned and signed bytes without promotion
• Assuming overflow without checking promotion
• Misinterpreting printf format specifiers

5. On an 8-bit MCU, you want to store the result of subtracting 150 from 100 using unsigned 8-bit variables:

unsigned char a = 100;
unsigned char b = 150;
unsigned char c = a - b;
printf("%u", c);

What will be printed and why?

hard

A. 206 because unsigned subtraction wraps around modulo 256

B. -50 because subtraction keeps sign in unsigned

C. 50 because it calculates absolute difference

D. Error due to negative result in unsigned

Solution

1. Step 1: Understand unsigned subtraction wrap-around

   Unsigned 8-bit subtraction wraps modulo 256. 100 - 150 = -50, which wraps to 256 - 50 = 206.

2. Step 2: Confirm printf output

   Variable c holds 206, so printf with %u prints 206.

3. Final Answer:

   206 because unsigned subtraction wraps around modulo 256 -> Option A

4. Quick Check:

   Unsigned subtraction wraps modulo 256 = 206 [OK]

Hint: Unsigned subtraction wraps modulo 256 on 8-bit MCU [OK]

Common Mistakes:

• Expecting negative numbers in unsigned arithmetic
• Assuming error on negative result
• Thinking subtraction gives absolute difference