Signed vs unsigned behavior on 8-bit MCU in Embedded C - Quick Revision & Key Differences

Recall & Review

beginner

What is the difference between signed and unsigned integers on an 8-bit MCU?

Signed integers can represent both positive and negative numbers, while unsigned integers represent only non-negative numbers (zero and positive). On an 8-bit MCU, signed integers typically use one bit for the sign, reducing the maximum positive value.

beginner

What is the range of an 8-bit signed integer?

An 8-bit signed integer ranges from -128 to 127. This is because one bit is used for the sign, and the remaining 7 bits represent the value.

beginner

What is the range of an 8-bit unsigned integer?

An 8-bit unsigned integer ranges from 0 to 255. All 8 bits are used to represent the value, so no bits are reserved for sign.

intermediate

What happens when you assign a negative value to an unsigned 8-bit variable on an MCU?

The negative value wraps around due to underflow, resulting in a large positive number. For example, assigning -1 to an unsigned 8-bit variable results in 255.

intermediate

How does overflow behave differently for signed and unsigned 8-bit integers?

For unsigned integers, overflow wraps around from the maximum value back to zero. For signed integers, overflow causes undefined behavior or wraps around in two's complement form, which can lead to negative values.

What is the maximum value of an 8-bit unsigned integer?

A 128

B 127

C 255

D 256

An 8-bit unsigned integer uses all 8 bits for value, so max is 2^8 - 1 = 255.

What is the range of an 8-bit signed integer?

A -128 to 127

B -256 to 255

C -127 to 128

D 0 to 255

Signed 8-bit integers use one bit for sign, so range is -128 to 127.

What happens if you assign -1 to an unsigned 8-bit variable?

A It stores 255

B It stores -1

C It causes an error

D It stores 0

Unsigned variables wrap around, so -1 becomes 255.

Which of these is true about overflow in unsigned 8-bit integers?

A It stays at max value

B It causes program crash

C It becomes negative

D It wraps around to zero

Unsigned overflow wraps around to zero due to modulo arithmetic.

Why is one bit reserved in signed 8-bit integers?

A To store parity

B To store the sign (positive or negative)

C To increase range

D To store overflow info

One bit is used to indicate if the number is positive or negative.

Explain the difference between signed and unsigned integers on an 8-bit MCU and how their ranges differ.

Think about how many values can be represented with 8 bits and how sign affects that.

Describe what happens when you assign a negative number to an unsigned 8-bit variable on an MCU.

Consider how unsigned variables handle values outside their range.

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