Writing an ISR (Interrupt Service Routine) in Embedded C - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare an ISR function for TIMER0 overflow.

Embedded C

void TIMER0_OVF_vect(void) [1] {
    // ISR code here
}

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

A__interrupt

Bvoid

CISR

Dinterrupt

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'void' instead of 'interrupt' keyword.

Forgetting to declare the function as an ISR.

The keyword 'interrupt' is used to declare an ISR function in embedded C for many compilers.

2fill in blank

medium

Complete the code to enable the TIMER0 overflow interrupt by setting the correct bit.

Embedded C

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

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

ATOIE1

BOCIE0

CTOIE0

DOCIE1

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing bits related to TIMER1 instead of TIMER0.

Using Output Compare Interrupt Enable bits instead of Overflow.

TOIE0 is the Timer/Counter0 Overflow Interrupt Enable bit.

3fill in blank

hard

Fix the error in the ISR declaration syntax.

Embedded C

void [1](void) interrupt 5 {
    // ISR code
}

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ATIMER0_OVF_vect

BISR_TIMER0

CTIMER1_OVF_vect

DISR

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect ISR function names.

Using generic names like 'ISR' without vector suffix.

The correct ISR name for TIMER0 overflow is 'TIMER0_OVF_vect' in many embedded C compilers.

4fill in blank

hard

Fill both blanks to correctly clear the interrupt flag inside the ISR.

Embedded C

void TIMER1_OVF_vect(void) {
    [1] &= ~(1 << [2]); // Clear interrupt flag
}

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

ATIFR

BTIMSK

CTOV1

DOCF1A

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to clear flags in TIMSK instead of TIFR.

Using wrong flag bits like OCF1A.

The interrupt flag for TIMER1 overflow is TOV1, and it is cleared by writing 0 to the bit in TIFR register.

5fill in blank

hard

Fill all three blanks to write an ISR that increments a counter and clears the interrupt flag.

Embedded C

volatile int counter = 0;

void [1](void) {
    counter[2];
    [3] &= ~(1 << TOV0);
}

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

ATIMER0_OVF_vect

B++

CTIFR

DTIMSK

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong ISR names.

Using += 1 instead of ++ (though valid, ++ is simpler here).

Clearing flags in TIMSK instead of TIFR.

The ISR name is TIMER0_OVF_vect, increment operator is ++, and the interrupt flag register is TIFR.

Practice

1. What is the main purpose of an ISR (Interrupt Service Routine) in embedded C?

easy

A. To allocate memory dynamically

B. To handle hardware events immediately when they occur

C. To run the main program loop continuously

D. To compile the program code

Solution

1. Step 1: Understand ISR role

   An ISR is a special function that runs automatically when a hardware event happens, like a button press or timer overflow.

2. Step 2: Compare options

   Only To handle hardware events immediately when they occur describes this immediate handling of hardware events. Other options describe unrelated tasks.

3. Final Answer:

   To handle hardware events immediately when they occur -> Option B

4. Quick Check:

   ISR = handle hardware events [OK]

Hint: ISRs respond instantly to hardware signals [OK]

Common Mistakes:

• Thinking ISR runs like normal functions
• Confusing ISR with main program loop
• Assuming ISR manages memory

2. Which of the following is the correct way to declare an ISR function in embedded C?

easy

A. void ISR_Handler() __interrupt { /* code */ }

B. void ISR_Handler() interrupt { /* code */ }

C. void ISR_Handler() __interrupt(0) { /* code */ }

D. void ISR_Handler() __isr { /* code */ }

Solution

1. Step 1: Recall ISR syntax in embedded C

   ISRs are declared with the __interrupt keyword followed by the interrupt vector number in parentheses.

2. Step 2: Check options

   void ISR_Handler() __interrupt(0) { /* code */ } uses __interrupt(0), which is the correct syntax. The others either miss the vector number, use the wrong keyword ('interrupt'), or a non-standard keyword ('__isr').

3. Final Answer:

   void ISR_Handler() __interrupt(0) { /* code */ } -> Option C

4. Quick Check:

   ISR syntax = __interrupt(vector) [OK]

Hint: Use __interrupt(vector_number) to declare ISR [OK]

Common Mistakes:

• Omitting interrupt vector number
• Using 'interrupt' instead of '__interrupt'
• Using non-standard keywords

3. Given the following ISR code snippet, what will be the value of flag after the interrupt occurs?

volatile int flag = 0;
void ISR_Timer() __interrupt(1) {
    flag = 1;
    // Interrupt flag cleared here
}

int main() {
    flag = 0;
    // Interrupt occurs here
    // ISR_Timer runs
    return flag;
}

medium

A. 1

B. 0

C. Undefined behavior

D. Compilation error

Solution

1. Step 1: Understand ISR effect on flag

   The ISR sets the variable 'flag' to 1 when the interrupt occurs.

2. Step 2: Check main function return

   Main sets flag to 0, then interrupt triggers ISR which sets flag to 1, so return value is 1.

3. Final Answer:

   1 -> Option A

4. Quick Check:

   ISR sets flag = 1, so return = 1 [OK]

Hint: ISR changes flag to 1 before main returns [OK]

Common Mistakes:

• Ignoring ISR effect on flag
• Assuming flag stays 0
• Confusing volatile usage

4. Identify the error in this ISR code snippet:

void ISR_UART() __interrupt(4) {
    if (UART_Interrupt_Flag) {
        // Handle UART interrupt
        UART_Interrupt_Flag = 0;
    }
}

medium

A. ISR should not clear interrupt flags inside it

B. ISR function must return an int value

C. ISR must be declared with 'interrupt' keyword, not '__interrupt'

D. Missing volatile keyword for UART_Interrupt_Flag variable

Solution

1. Step 1: Check variable usage in ISR

   Interrupt flags accessed in ISR should be declared volatile to prevent compiler optimization issues.

2. Step 2: Analyze other options

   Clearing flags inside ISR is correct practice; __interrupt is correct keyword; ISR functions return void, not int.

3. Final Answer:

   Missing volatile keyword for UART_Interrupt_Flag variable -> Option D

4. Quick Check:

   Interrupt flags must be volatile [OK]

Hint: Declare interrupt flags volatile to avoid optimization bugs [OK]

Common Mistakes:

• Not declaring flags volatile
• Wrong ISR declaration keyword
• Expecting ISR to return a value

5. You want to write an ISR for a timer interrupt that increments a counter and clears the interrupt flag. Which of the following is the best practice to ensure correct ISR behavior?

hard

A. Increment counter, then clear interrupt flag inside ISR

B. Clear interrupt flag, then increment counter inside ISR

C. Increment counter outside ISR and clear flag inside ISR

D. Do not clear interrupt flag; let main program clear it

Solution

1. Step 1: Understand ISR order for flag and counter

   Best practice is to handle the event (increment counter) first, then clear the interrupt flag to acknowledge the interrupt.

2. Step 2: Evaluate options

   Increment counter, then clear interrupt flag inside ISR follows this order. Clear interrupt flag, then increment counter inside ISR clears flag before handling event, which may cause missed interrupts. Increment counter outside ISR and clear flag inside ISR splits logic incorrectly. Do not clear interrupt flag; let main program clear it risks repeated interrupts.

3. Final Answer:

   Increment counter, then clear interrupt flag inside ISR -> Option A

4. Quick Check:

   Handle event first, then clear flag [OK]

Hint: Process event before clearing interrupt flag [OK]

Common Mistakes:

• Clearing flag before handling event
• Leaving flag uncleared causing repeated interrupts
• Handling event outside ISR