Direction register vs data register in Embedded C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set pin 0 of PORTB as output.

Embedded C

DDRB |= (1 << [1]);

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

A1

B2

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using PORTB instead of DDRB to set direction.

Setting the wrong bit number.

Setting bit 0 in DDRB configures pin 0 of PORTB as output.

2fill in blank

medium

Complete the code to write a high value to pin 2 of PORTC.

Embedded C

PORTC |= (1 << [1]);

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

A2

B1

C3

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using DDRC instead of PORTC to write data.

Setting the wrong bit number.

Setting bit 2 in PORTC writes a high value to pin 2.

3fill in blank

hard

Fix the error in the code to configure pin 1 of PORTD as input.

Embedded C

DDRD [1]= ~(1 << 1);

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

A|

B&

C^

D=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|=' instead of '&=' to clear the bit.

Using '=' which overwrites all bits.

Using '&=' with the negated bit clears bit 1, setting pin 1 as input.

4fill in blank

hard

Fill both blanks to set pin 3 of PORTB as output and write it low.

Embedded C

DDRB [1]= (1 << 3);
PORTB [2]= ~(1 << 3);

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

A|

B&

C=

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|=' or '&=' which only modifies bits, not sets exact value.

Not negating the bit when writing low.

Using '=' sets DDRB bit 3 as output by overwriting the register, and '&=' with negation clears PORTB bit 3 to write it low.

5fill in blank

hard

Fill all three blanks to configure pin 4 of PORTC as output, set it high, and then clear it.

Embedded C

DDRC [1]= (1 << 4);
PORTC [2]= (1 << 4);
PORTC [3]= ~(1 << 4);

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

A|

B&

C=

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|=' or '&=' which only modifies bits without overwriting.

Not negating the bit mask when clearing the pin.

Using '=' assigns exact values to DDRC and PORTC to set pin 4 output and set it high, then '&=' with negation clears the pin.

Practice

1. What is the primary role of a direction register in embedded C programming?

easy

A. It resets the microcontroller.

B. It stores the data value to be sent through the pin.

C. It sets whether a pin is an input or an output.

D. It controls the clock speed of the microcontroller.

Solution

1. Step 1: Understand the function of direction register

   The direction register configures each pin as either input or output.

2. Step 2: Differentiate from data register

   The data register holds the actual value sent or received, not the pin mode.

3. Final Answer:

   It sets whether a pin is an input or an output. -> Option C

4. Quick Check:

   Direction register = Pin mode control [OK]

Hint: Direction register controls pin mode, not data value [OK]

Common Mistakes:

• Confusing direction register with data register
• Thinking direction register stores data values
• Mixing up pin configuration with clock control

2. Which of the following is the correct way to set pin 3 as output using a direction register named DIR_REG in embedded C?

easy

A. DIR_REG = 1 << 3;

B. DIR_REG |= (1 << 3);

C. DIR_REG &= ~(1 << 3);

D. DIR_REG ^= (1 << 3);

Solution

1. Step 1: Understand setting a bit without affecting others

   Using |= with (1 << 3) sets bit 3 to 1, keeping other bits unchanged.

2. Step 2: Analyze other options

   DIR_REG = 1 << 3; overwrites all bits, DIR_REG &= ~(1 << 3); clears bit 3, DIR_REG ^= (1 << 3); toggles bit 3.

3. Final Answer:

   DIR_REG |= (1 << 3); -> Option B

4. Quick Check:

   Use |= to set bits safely [OK]

Hint: Use |= to set a bit without changing others [OK]

Common Mistakes:

• Using = instead of |= which overwrites all bits
• Using &=~ which clears the bit instead of setting
• Using ^= which toggles the bit unpredictably

3. Given the following code snippet, what is the final value of DATA_REG if initially it was 0x00?

DIR_REG |= (1 << 2);  // Set pin 2 as output
DATA_REG |= (1 << 2); // Set pin 2 high
DATA_REG &= ~(1 << 1); // Clear pin 1
DATA_REG ^= (1 << 2);  // Toggle pin 2

medium

A. 0x04

B. 0x06

C. 0x02

D. 0x00

Solution

1. Step 1: Trace DATA_REG changes step-by-step

   Initially DATA_REG = 0x00.
   After DATA_REG |= (1 << 2); DATA_REG = 0x04.
   After DATA_REG &= ~(1 << 1); bit 1 cleared, no change, DATA_REG = 0x04.
   After DATA_REG ^= (1 << 2); toggles bit 2, DATA_REG = 0x00.

2. Step 2: Confirm final value

   Final DATA_REG value is 0x00 after toggling bit 2 off.

3. Final Answer:

   0x00 -> Option D

4. Quick Check:

   Toggle bit 2 twice returns to 0 [OK]

Hint: Toggle a bit twice returns it to original state [OK]

Common Mistakes:

• Ignoring toggle effect and assuming bit stays set
• Confusing bit positions (bit 1 vs bit 2)
• Not tracking bit clearing correctly

4. Identify the error in the following code that tries to set pin 0 as output and set it high:

DIR_REG = DIR_REG & (1 << 0);
DATA_REG = DATA_REG | (1 << 0);

medium

A. Using & instead of | to set direction bit clears other bits.

B. DATA_REG should use &= instead of | to set pin high.

C. DIR_REG should be assigned 0, not bitwise operation.

D. No error, code works correctly.

Solution

1. Step 1: Analyze DIR_REG operation

   DIR_REG & (1 << 0) clears all bits except bit 0, overwriting other pin settings.

2. Step 2: Correct way to set bit without clearing others

   Use DIR_REG |= (1 << 0) to set bit 0 without affecting others.

3. Final Answer:

   Using & instead of | to set direction bit clears other bits. -> Option A

4. Quick Check:

   Use |= to set bits, not & [OK]

Hint: Use |= to set bits, & clears bits [OK]

Common Mistakes:

• Using & instead of |= to set bits
• Thinking |= clears bits
• Assuming DATA_REG needs &= to set bits

5. You want to configure pins 0, 2, and 4 as outputs and set pins 0 and 4 high without changing other pins. Which sequence of operations correctly achieves this?

// Initial values: DIR_REG = 0x08, DATA_REG = 0x02

hard

A. DIR_REG |= (1 << 0) | (1 << 2) | (1 << 4);
DATA_REG |= (1 << 0) | (1 << 4);

B. DIR_REG = (1 << 0) | (1 << 2) | (1 << 4);
DATA_REG = (1 << 0) | (1 << 4);

C. DIR_REG &= ~((1 << 1) | (1 << 3));
DATA_REG |= (1 << 0) | (1 << 4);

D. DIR_REG |= (1 << 1) | (1 << 3);
DATA_REG |= (1 << 2);

Solution

1. Step 1: Set pins 0, 2, 4 as outputs without clearing others

   Use DIR_REG |= (1 << 0) | (1 << 2) | (1 << 4); to set these bits safely.

2. Step 2: Set pins 0 and 4 high without affecting others

   Use DATA_REG |= (1 << 0) | (1 << 4); to set those pins high.

3. Step 3: Analyze other options

   DIR_REG = (1 << 0) | (1 << 2) | (1 << 4); DATA_REG = (1 << 0) | (1 << 4); overwrites all bits, clearing others.
   DIR_REG &= ~((1 << 1) | (1 << 3)); DATA_REG |= (1 << 0) | (1 << 4); clears pins 1 and 3 but does not set required pins.
   DIR_REG |= (1 << 1) | (1 << 3); DATA_REG |= (1 << 2); sets wrong pins.

4. Final Answer:

   DIR_REG |= (1 << 0) | (1 << 2) | (1 << 4); DATA_REG |= (1 << 0) | (1 << 4); -> Option A

5. Quick Check:

   Use |= to set multiple bits safely [OK]

Hint: Use |= with combined bits to set multiple pins safely [OK]

Common Mistakes:

• Using = which overwrites other pins
• Setting wrong pins by mistake
• Clearing pins unintentionally