Why timers are needed in Embedded C - Test Your Understanding

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the timer.

Embedded C

TIMER_START([1]);

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

Adelay

Btimer1

Cstart

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using action words like 'start' instead of the timer name.

Passing delay or count instead of the timer identifier.

The function TIMER_START needs the timer name to start it. Here, timer1 is the correct timer identifier.

2fill in blank

medium

Complete the code to check if the timer has expired.

Embedded C

if ([1]_expired()) {
    // Timer expired actions
}

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

Atimer2

Btimer_expired

Ctimer1

Dexpired

Attempts:

3 left

💡 Hint

Common Mistakes

Using generic words like 'expired' without the timer name.

Using incorrect timer names like 'timer2' if not defined.

The function to check if a timer expired is usually named with the timer's name followed by _expired(). Here, timer1_expired() is correct.

3fill in blank

hard

Fix the error in the timer initialization code.

Embedded C

void init_timer() {
    [1] = 0;
    TIMER_CONFIGURE(timer1, 1000);
}

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

Atimer

Btimer1

Ccount

Dtimer1_count

Attempts:

3 left

💡 Hint

Common Mistakes

Resetting the timer variable instead of the count variable.

Using generic names like 'count' without context.

The variable timer1_count should be reset to zero to initialize the timer count properly.

4fill in blank

hard

Fill both blanks to create a delay using the timer.

Embedded C

while (![1]_expired()) {
    [2]();
}

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

Atimer3

Bdo_nothing

Ctimer1

Dwait

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong timer names in the expiration check.

Leaving the loop body empty causing high CPU usage.

The loop waits until timer1_expired() returns true. Inside the loop, do_nothing() is called to avoid busy waiting.

5fill in blank

hard

Fill all three blanks to set up and start a timer with a callback.

Embedded C

void setup() {
    [1] = 500;
    TIMER_CONFIGURE(timer2, [2]);
    TIMER_START([3]);
}

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

Adelay_ms

Ctimer2

Dtimer1

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for delay in the first two blanks.

Starting the wrong timer.

The variable delay_ms is set to 500 milliseconds. Then TIMER_CONFIGURE uses delay_ms to set the timer period. Finally, TIMER_START(timer2) starts the timer.

Practice

1. Why do embedded C programs often use timers?

easy

A. To change the program language automatically

B. To make the program run faster

C. To measure time intervals without stopping the program

D. To increase memory size

Solution

1. Step 1: Understand the role of timers in embedded systems

   Timers allow the program to keep track of time without pausing or stopping other tasks.

2. Step 2: Compare other options

   Options A, B, and C do not relate to timing or program control.

3. Final Answer:

   To measure time intervals without stopping the program -> Option C

4. Quick Check:

   Timers track time without stopping = D [OK]

Hint: Timers track time without pausing program flow [OK]

Common Mistakes:

• Thinking timers speed up the program
• Confusing timers with memory management
• Believing timers change programming language

2. Which of the following is the correct way to start a timer in embedded C?

easy

A. timer_start();

B. start_timer();

C. TimerStart();

D. timer.begin();

Solution

1. Step 1: Recognize common embedded C timer function naming

   Functions in embedded C often use lowercase with underscores, like timer_start().

2. Step 2: Check other options for syntax

   start_timer() and TimerStart() are possible but less common; timer.begin() looks like C++ or Arduino style.

3. Final Answer:

   timer_start(); -> Option A

4. Quick Check:

   Embedded C uses lowercase with underscores = A [OK]

Hint: Embedded C uses lowercase_underscore style functions [OK]

Common Mistakes:

• Using C++ style method calls
• Capitalizing function names incorrectly
• Mixing Arduino style with embedded C

3. What will be the output of this embedded C code snippet?

volatile int flag = 0;
void timer_interrupt() {
  flag = 1;
}
int main() {
  while(flag == 0) {
    // wait
  }
  printf("Timer done\n");
  return 0;
}

medium

A. Timer done

B. Runtime error

C. Compilation error

D. No output, program stuck in loop

Solution

1. Step 1: Analyze the flag variable and interrupt

   The flag is set to 1 only inside timer_interrupt(), which is not called in main.

2. Step 2: Understand program flow

   Main waits in while(flag == 0) forever because flag never changes.

3. Final Answer:

   No output, program stuck in loop -> Option D

4. Quick Check:

   Interrupt not triggered = infinite wait = B [OK]

Hint: Interrupt must be triggered to change flag [OK]

Common Mistakes:

• Assuming interrupt runs automatically
• Expecting output without interrupt call
• Ignoring volatile keyword purpose

4. Identify the error in this timer usage code snippet:

int timer_flag = 0;
void timer_ISR() {
  timer_flag = 1;
}
int main() {
  while(timer_flag == 0) {
    // wait
  }
  printf("Timer expired\n");
  return 0;
}

medium

A. printf() cannot be used in embedded C

B. timer_flag should be declared volatile

C. timer_ISR() should return int

D. while loop condition is incorrect

Solution

1. Step 1: Check variable shared with interrupt

   timer_flag is changed in ISR and read in main, so it must be volatile to prevent optimization.

2. Step 2: Review other options

   ISR return type can be void; printf() can be used if supported; while condition is logically correct.

3. Final Answer:

   timer_flag should be declared volatile -> Option B

4. Quick Check:

   Shared variable with ISR needs volatile = C [OK]

Hint: Use volatile for variables changed by interrupts [OK]

Common Mistakes:

• Ignoring volatile keyword
• Changing ISR return type unnecessarily
• Thinking printf is always disallowed

5. You want to blink an LED every 1 second without stopping other tasks. Which timer approach is best in embedded C?

hard

A. Use a timer interrupt to toggle LED flag every 1 second

B. Use a blocking delay loop for 1 second

C. Use a recursive function to blink LED

D. Use a global variable incremented in main loop

Solution

1. Step 1: Understand blinking without stopping other tasks

   Blocking delay (Use a blocking delay loop for 1 second) stops other tasks; recursive function (Use a recursive function to blink LED) is inefficient; global variable incremented in main (Use a global variable incremented in main loop) may not be precise.

2. Step 2: Use timer interrupt for precise, non-blocking control

   Timer interrupt toggles LED flag every 1 second without blocking main program.

3. Final Answer:

   Use a timer interrupt to toggle LED flag every 1 second -> Option A

4. Quick Check:

   Timer interrupt enables non-blocking blinking = A [OK]

Hint: Use timer interrupts for non-blocking periodic tasks [OK]

Common Mistakes:

• Using blocking delays that freeze program
• Trying recursion for timing
• Relying on main loop variable without timer