Start and stop conditions in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this I2C start condition code snippet?

Consider the following embedded C code that attempts to generate an I2C start condition. What will be the value of the variable status after execution?

Embedded C

volatile int SDA = 1; // Data line
volatile int SCL = 1; // Clock line
int status = 0;

// Generate start condition
SDA = 1;
SCL = 1;
if (SDA == 1 && SCL == 1) {
    SDA = 0; // Pull SDA low while SCL high
    if (SDA == 0 && SCL == 1) {
        status = 1; // Start condition successful
    } else {
        status = -1; // Failed
    }
} else {
    status = -2; // Bus not idle
}

A status = 1

B status = -1

C status = -2

D status = 0

💡 Hint

Remember, the I2C start condition is when SDA goes low while SCL is high.

The code first ensures both SDA and SCL are high (bus idle). Then it pulls SDA low while SCL remains high, which is the start condition. So status is set to 1.

❓ Predict Output

intermediate

What does this code print when generating an I2C stop condition?

This embedded C code tries to generate an I2C stop condition. What will be printed?

Embedded C

#include <stdio.h>

volatile int SDA = 0; // Data line
volatile int SCL = 1; // Clock line

int main() {
    // Generate stop condition
    SDA = 0;
    SCL = 1;
    SDA = 1; // Release SDA while SCL high

    if (SDA == 1 && SCL == 1) {
        printf("STOP_OK\n");
    } else {
        printf("STOP_FAIL\n");
    }
    return 0;
}

A Compilation error

B STOP_FAIL

C No output

D STOP_OK

💡 Hint

The stop condition is when SDA goes from low to high while SCL is high.

The code sets SDA low, then SCL high, then SDA high while SCL is high. This matches the stop condition, so it prints STOP_OK.

🔧 Debug

advanced

Why does this I2C start condition code fail to detect the start?

This code is intended to detect an I2C start condition but incorrectly sets start_detected to 1. What is the bug?

Embedded C

volatile int SDA = 1;
volatile int SCL = 1;
int start_detected = 0;

void detect_start() {
    if (SDA == 1 && SCL == 1) {
        SDA = 0; // Pull SDA low
    }
    if (SDA == 0 && SCL == 1) {
        start_detected = 1;
    } else {
        start_detected = 0;
    }
}

int main() {
    detect_start();
    return start_detected;
}

A start_detected should be initialized to 1, not 0.

B The code changes SDA before checking the condition, so the first if is always true and start_detected is never set correctly.

C SCL should be pulled low before SDA to detect start condition.

D The function detect_start() never returns a value, causing undefined behavior.

💡 Hint

Think about the order of operations and when the condition is checked.

The code sets SDA to 0 inside the first if, then immediately checks if SDA is 0 and SCL is 1. Because SDA was just changed, the condition is always true, but the logic flow causes start_detected to be set incorrectly. The detection logic should check the condition before changing SDA.

📝 Syntax

advanced

Which option causes a syntax error in this I2C stop condition code?

Given the following code snippet, which option will cause a syntax error when compiling?

Embedded C

volatile int SDA = 0;
volatile int SCL = 1;

void generate_stop() {
    SDA = 0;
    SCL = 1;
    SDA = 1; // Release SDA while SCL high
}

A void generate_stop() { SDA = 0; SCL = 1 SDA = 1; }

B void generate_stop() { SDA = 0; SCL = 1; SDA = 1; }

C void generate_stop() { SDA = 0; SCL = 1; SDA = 1; return; }

D } ;1 = ADS ;1 = LCS ;0 = ADS { )(pots_etareneg diov

💡 Hint

Look carefully for missing semicolons.

Option A is missing a semicolon between 'SCL = 1' and 'SDA = 1', causing a syntax error. The others are syntactically correct.

🚀 Application

expert

How many times does the start condition get detected in this signal sequence?

Given the following signal changes over time, how many start conditions are detected? A start condition is when SDA goes from high to low while SCL is high. Time steps (SDA, SCL): 1: (1,1) 2: (0,1) 3: (0,0) 4: (1,0) 5: (1,1) 6: (0,1) 7: (0,1) 8: (1,1) Assume the detection code checks transitions at each time step.

A 1

B 3

C 2

D 0

💡 Hint

Look for moments when SDA changes from 1 to 0 while SCL is 1.

Start conditions occur at time steps 2 (SDA 1->0 while SCL=1) and 6 (SDA 1->0 while SCL=1). So total 2 start conditions.