Double buffer technique in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of double buffer swap simulation

What is the output of this code simulating a double buffer swap in embedded C?

Embedded C

#include <stdio.h>

int main() {
    int buffer1 = 10;
    int buffer2 = 20;
    int *frontBuffer = &buffer1;
    int *backBuffer = &buffer2;

    // Swap buffers
    int *temp = frontBuffer;
    frontBuffer = backBuffer;
    backBuffer = temp;

    printf("Front buffer value: %d\n", *frontBuffer);
    printf("Back buffer value: %d\n", *backBuffer);
    return 0;
}

A

Front buffer value: 0
Back buffer value: 0

B

Front buffer value: 10
Back buffer value: 20

C Compilation error due to pointer swap

D

Front buffer value: 20
Back buffer value: 10

💡 Hint

Remember that swapping pointers changes which buffer is front and which is back.

The pointers frontBuffer and backBuffer are swapped, so frontBuffer points to buffer2 (20) and backBuffer points to buffer1 (10).

🧠 Conceptual

intermediate

Purpose of double buffering in embedded systems

Why is double buffering commonly used in embedded systems?

A To allow simultaneous reading and writing without data corruption

B To reduce the size of the program memory

C To increase the clock speed of the processor

D To simplify the hardware design by using only one buffer

💡 Hint

Think about how data is accessed and updated during processing.

Double buffering allows one buffer to be displayed or read while the other is being written to, preventing data corruption and flickering.

🔧 Debug

advanced

Identify the error in double buffer swap code

What error does this code produce when swapping double buffers?

Embedded C

int buffer1 = 5;
int buffer2 = 15;
int *frontBuffer = &buffer1;
int *backBuffer = &buffer2;

// Incorrect swap
frontBuffer = backBuffer;
backBuffer = frontBuffer;

printf("Front: %d, Back: %d\n", *frontBuffer, *backBuffer);

A Both pointers point to the same buffer causing data overwrite

B Compilation error due to missing semicolon

C Runtime segmentation fault

D Correct output: Front: 15, Back: 5

💡 Hint

Check what happens after the first assignment to frontBuffer.

The swap is done incorrectly: after frontBuffer = backBuffer, both pointers point to buffer2, so backBuffer = frontBuffer does not restore the original pointer.

📝 Syntax

advanced

Identify syntax error in double buffer initialization

Which option contains the correct syntax to declare and initialize two buffers and pointers for double buffering?

A

int buffer1, buffer2;
int *frontBuffer = buffer1;
int *backBuffer = buffer2;

B

int buffer1 = 0, buffer2 = 0;
int *frontBuffer = &buffer1;
int *backBuffer = &buffer2;

C

int buffer1 = 0; int buffer2 = 0
int *frontBuffer = &buffer1
int *backBuffer = &buffer2

D

int buffer1 = 0, buffer2 = 0;
int frontBuffer = &buffer1;
int backBuffer = &buffer2;

💡 Hint

Remember pointers need to be assigned addresses with & and statements end with semicolons.

Option B correctly declares buffers and pointers with proper syntax and semicolons. Option B assigns pointers to values, not addresses. Option B misses semicolons. Option B declares pointers as int, not int*.

🚀 Application

expert

Number of buffer swaps in a frame rendering loop

In an embedded system using double buffering, a frame rendering loop runs 60 times per second. How many buffer swaps occur after 10 seconds?

A 1200

B 60

C 600

D 30

💡 Hint

Each frame requires one buffer swap.

At 60 frames per second, each frame swap happens once. Over 10 seconds, 60 * 10 = 600 swaps occur.