Signed vs unsigned behavior on 8-bit MCU in Embedded C - Visual Side-by-Side Comparison

Concept Flow - Signed vs unsigned behavior on 8-bit MCU

Start with 8-bit variable

↓

Assign value (signed or unsigned)

↓

Store in 8 bits

↓

Interpret bits as signed or unsigned

↓

Perform arithmetic or comparison

↓

Result depends on signedness

↓

Use result in program

This flow shows how an 8-bit variable stores values and how interpreting bits as signed or unsigned changes the result.

Execution Sample

Embedded C

unsigned char u = 250;
signed char s = 250;

printf("u = %u\n", u);
printf("s = %d\n", s);

This code assigns 250 to both unsigned and signed 8-bit variables and prints their values.

Execution Table

Step	Variable	Value Assigned	Stored Bits (8-bit)	Interpreted Value	Print Output
1	u (unsigned char)	250	11111010	250	u = 250
2	s (signed char)	250	11111010	-6	s = -6
3	End	-	-	-	Execution ends

💡 Execution stops after printing both values.

Variable Tracker

Variable	Start	After Assignment	Final
u (unsigned char)	undefined	250	250
s (signed char)	undefined	250 (stored as 11111010)	-6 (interpreted signed)

Key Moments - 3 Insights

Why does assigning 250 to a signed 8-bit variable show -6 when printed?

Because 8-bit signed variables use two's complement. The bits 11111010 represent -6 in signed form, as shown in execution_table row 2.

Why does the unsigned variable print 250 correctly?

Unsigned variables interpret bits as positive numbers only, so 11111010 is 250 decimal, as seen in execution_table row 1.

What happens if arithmetic is done on signed vs unsigned variables?

Arithmetic results depend on interpretation: signed variables can represent negative numbers, unsigned cannot, affecting overflow and comparisons.

Visual Quiz - 3 Questions

Look at the execution_table, what is the interpreted value of the signed char variable after assignment?

A 250

B -6

C 6

D -250

💡 Hint

Check execution_table row 2 under 'Interpreted Value' column.

At which step does the unsigned variable get assigned its value?

A Step 2

B Step 3

C Step 1

D No assignment

💡 Hint

Look at execution_table row 1 under 'Variable' and 'Value Assigned'.

If the signed char variable was assigned 130 instead of 250, what would be its interpreted value?

A -126

B 130

C -130

D 126

💡 Hint

Signed 8-bit range is -128 to 127; 130 stored as bits wraps to negative number similar to 250 case.

Concept Snapshot

Signed vs unsigned 8-bit variables store the same bits but interpret them differently.
Unsigned chars hold 0 to 255.
Signed chars hold -128 to 127 using two's complement.
Assigning values >127 to signed chars wraps to negative numbers.
Printing must match variable type to show correct value.

Full Transcript

This lesson shows how 8-bit microcontrollers store values in variables. When you assign a number like 250 to an unsigned 8-bit variable, it stores the bits 11111010 and prints 250. But if you assign 250 to a signed 8-bit variable, it stores the same bits but interprets them as -6 because signed 8-bit uses two's complement. This difference affects how arithmetic and comparisons work. Always remember to use the correct type and print format to see the expected values.

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