Nested interrupt handling in FreeRTOS - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of nested interrupt priority handling

Consider a FreeRTOS system with two interrupts: ISR1 with priority 3 and ISR2 with priority 2 (lower number means higher priority). ISR1 triggers ISR2. What will be the output sequence?

FreeRTOS

#include "FreeRTOS.h"
#include "task.h"
#include <stdio.h>

volatile int output = 0;

void ISR2_Handler(void) {
    output = output * 10 + 2;
}

void ISR1_Handler(void) {
    output = output * 10 + 1;
    // Simulate triggering ISR2
    ISR2_Handler();
    output = output * 10 + 3;
}

int main() {
    output = 0;
    ISR1_Handler();
    printf("%d", output);
    return 0;
}

A 123

B 132

C 12

D 321

💡 Hint

Think about the order in which the interrupts execute and how nested interrupts affect the output.

ISR1 starts and appends '1'. Then it triggers ISR2 which appends '2'. After ISR2 finishes, ISR1 resumes and appends '3'. So the output is '123'.

🧠 Conceptual

intermediate

Which interrupt can preempt in nested handling?

In FreeRTOS, if two interrupts have priorities 5 and 7 (lower number means higher priority), which interrupt can preempt the other during nested interrupt handling?

A Interrupt with priority 7 can preempt priority 5

B Interrupt with priority 5 can preempt priority 7

C Neither can preempt each other

D Both can preempt each other simultaneously

💡 Hint

Remember that lower priority number means higher priority in FreeRTOS.

In FreeRTOS, lower priority number means higher priority. So priority 5 is higher than 7 and can preempt it.

🔧 Debug

advanced

Identify the error in nested interrupt enabling

What error will occur with this nested interrupt code snippet in FreeRTOS?

FreeRTOS

void ISR1_Handler(void) {
    portDISABLE_INTERRUPTS();
    // Critical section
    portENABLE_INTERRUPTS();
    ISR2_Handler();
}

void ISR2_Handler(void) {
    // Nested interrupt code
}

A Compilation error due to missing semicolon

B Nested interrupt will run correctly without issues

C Stack overflow due to recursive interrupts

D ISR2 will never run because interrupts are disabled

💡 Hint

Check when interrupts are enabled or disabled during nested calls.

Interrupts are disabled before ISR2 is called, so ISR2 cannot run as a nested interrupt.

📝 Syntax

advanced

Identify syntax error in nested interrupt declaration

Which option contains a syntax error in declaring nested interrupt handlers in FreeRTOS?

FreeRTOS

void ISR1_Handler(void) __attribute__((interrupt));
void ISR2_Handler(void) __attribute__((interrupt));

A void ISR1_Handler(void) __attribute__((interrupt));

B void ISR1_Handler() __attribute__((interrupt));

C void ISR2_Handler(void) __attribute__((interrupt) {});

D void ISR2_Handler(void) __attribute__((interrupt));

💡 Hint

Look carefully at the attribute syntax and function declaration.

Option C has incorrect syntax: the attribute is followed by braces which is invalid.

🚀 Application

expert

Calculate maximum nested interrupt depth

Given a FreeRTOS system with 8 interrupt priority levels (0 highest, 7 lowest), and nested interrupts allowed only if the new interrupt has higher priority, what is the maximum possible nested interrupt depth starting from priority 7?

A 7

B 8

C 1

D 0

💡 Hint

Think about how many higher priority interrupts exist above priority 7.

Starting from priority 7, interrupts with priorities 6,5,4,3,2,1,0 can preempt, so maximum nested depth is 7.