GPIO register configuration in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this GPIO configuration code?

Consider the following embedded C code snippet configuring a GPIO port. What will be the value of GPIOA->MODER after execution?

Embedded C

typedef struct {
    volatile unsigned int MODER;
} GPIO_TypeDef;

GPIO_TypeDef GPIOA_instance = {0};
GPIO_TypeDef *GPIOA = &GPIOA_instance;

int main() {
    GPIOA->MODER = 0;
    GPIOA->MODER |= (1 << (2 * 3)); // Set pin 3 to output mode (01)
    GPIOA->MODER |= (2 << (2 * 4)); // Set pin 4 to alternate function mode (10)
    return GPIOA->MODER;
}

A 0x00000240

B 0x00000140

C 0x00000400

D 0x00000340

💡 Hint

Remember each pin uses 2 bits in MODER. Pin 3 bits are bits 6-7, pin 4 bits are bits 8-9.

Pin 3 set to output mode means bits 6-7 = 01 (1 << 6 = 0x40). Pin 4 set to alternate function mode means bits 8-9 = 10 (2 << 8 = 0x200). Total: 0x240.

🧠 Conceptual

intermediate

How many bits control the mode of a single GPIO pin?

In a typical microcontroller GPIO port configuration register, how many bits are used to set the mode (input, output, alternate function, analog) of one pin?

A 1 bit

B 4 bits

C 8 bits

D 2 bits

💡 Hint

Think about how many modes are possible and how many bits are needed to represent them.

Each GPIO pin mode is selected by 2 bits because there are 4 possible modes: 00=input, 01=output, 10=alternate function, 11=analog.

🔧 Debug

advanced

Why does this GPIO output configuration fail to set pin 3 as output?

Examine the code below. It attempts to set GPIO pin 3 as output mode but does not work as expected. What is the cause?

Embedded C

GPIOA->MODER &= ~(0x3 << 3);
GPIOA->MODER |= (0x1 << 3);

A Bit shifts are incorrect; should shift by 6 and 6 respectively

B Masking uses wrong bits; should use 0x1 instead of 0x3

C The code sets pin 1 instead of pin 3

D No error; code works correctly

💡 Hint

Remember each pin uses 2 bits in MODER. Check the shift amount carefully.

Each pin uses 2 bits in MODER. Pin 3's bits are bits 6 and 7 (2*3=6). The code shifts by 3 which affects bits 3 and 4, not pin 3. Correct shift is 6.

📝 Syntax

advanced

What error does this GPIO register configuration code produce?

Consider this code snippet. What error will it produce?

Embedded C

GPIOA->MODER = 0;
GPIOA->MODER |= 1 << 2 * 5;

A No error; sets pin 5 mode bit 2 correctly

B Logical error; shifts 1 by 2 then multiplies by 5, wrong bit set

C SyntaxError due to missing parentheses around shift expression

D Runtime error due to invalid pointer

💡 Hint

Operator precedence matters in bit shifts and multiplication.

Without parentheses, the expression is evaluated as (1 << 2) * 5 = 4 * 5 = 20, so 20 is ORed, not the intended 1 << 10. This causes wrong bits to be set.

🚀 Application

expert

Calculate the final MODER register value after multiple pin configurations

Given the following code configuring GPIO pins, what is the final hexadecimal value of GPIOB->MODER?

Embedded C

typedef struct {
    volatile unsigned int MODER;
} GPIO_TypeDef;

GPIO_TypeDef GPIOB_instance = {0};
GPIO_TypeDef *GPIOB = &GPIOB_instance;

int main() {
    GPIOB->MODER = 0;
    // Set pin 0 to analog mode (11)
    GPIOB->MODER |= (3 << (2 * 0));
    // Set pin 1 to output mode (01)
    GPIOB->MODER |= (1 << (2 * 1));
    // Set pin 2 to alternate function mode (10)
    GPIOB->MODER |= (2 << (2 * 2));
    // Set pin 3 to input mode (00)
    GPIOB->MODER &= ~(3 << (2 * 3));
    return GPIOB->MODER;
}

A 0x00000023

B 0x0000002B

C 0x00000027

D 0x0000003B

💡 Hint

Calculate each pin's bits carefully and apply the clearing operations in order.

Pin 0 analog (11): 3 << 0 = 0x3
Pin 1 output (01): 1 << 2 = 0x4 (total 0x7)
Pin 2 alt (10): 2 << 4 = 0x20 (total 0x27)
Pin 3 input (00): clears bits 6-7 (already 0), no change.
Final: 0x27