Digital input (GPIO.input) in Raspberry Pi - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Reading a button press state

What is the output of this Raspberry Pi GPIO code when the button connected to GPIO pin 17 is pressed?

Raspberry Pi

import RPi.GPIO as GPIO
GPIO.setmode(GPIO.BCM)
GPIO.setup(17, GPIO.IN, pull_up_down=GPIO.PUD_UP)

button_state = GPIO.input(17)
print(button_state)

A Raises RuntimeError

B 1

C 0

D None

💡 Hint

Remember that the button is connected with a pull-up resistor, so pressing it connects the pin to ground.

With a pull-up resistor, the input reads 1 when the button is not pressed and 0 when pressed because pressing connects the pin to ground.

❓ Predict Output

intermediate

Detecting button release state

What will be printed by this code snippet if the button connected to GPIO pin 22 is not pressed?

Raspberry Pi

import RPi.GPIO as GPIO
GPIO.setmode(GPIO.BCM)
GPIO.setup(22, GPIO.IN, pull_up_down=GPIO.PUD_DOWN)

state = GPIO.input(22)
print(state)

A 0

B 1

C None

D Raises RuntimeError

💡 Hint

Think about what the pull-down resistor does when the button is not pressed.

A pull-down resistor keeps the pin at LOW (0) when the button is not pressed, so GPIO.input returns 0.

🔧 Debug

advanced

Why does this GPIO.input code always print 1?

Consider this code snippet. The button is connected to GPIO pin 5 with a pull-up resistor. The code always prints 1, even when the button is pressed. What is the most likely cause?

Raspberry Pi

import RPi.GPIO as GPIO
GPIO.setmode(GPIO.BCM)
GPIO.setup(5, GPIO.IN)

print(GPIO.input(5))

A GPIO.input always returns 1 by default.

B The button is wired incorrectly, so it never connects to ground.

C The pull-up resistor is missing in setup, so the pin floats and reads 1 randomly.

D The code is missing pull_up_down=GPIO.PUD_UP in setup, so the pin floats and reads 1.

💡 Hint

Think about what happens if you don't specify pull_up_down in GPIO.setup for an input pin.

If pull_up_down is not set, the pin is floating and may read random values. In this case, it reads 1 because of electrical noise or internal state.

🧠 Conceptual

advanced

Understanding GPIO.input behavior with pull resistors

Which statement correctly describes the behavior of GPIO.input(pin) when a button is connected with a pull-down resistor?

A GPIO.input(pin) returns 0 when the button is pressed because the pin is connected to ground through the button.

B GPIO.input(pin) returns 1 when the button is pressed because the pin is connected to 3.3V through the button.

C GPIO.input(pin) returns 1 when the button is not pressed because the pull-down resistor pulls the pin high.

D GPIO.input(pin) returns None if the button is pressed.

💡 Hint

Think about what a pull-down resistor does and what pressing the button does to the pin voltage.

A pull-down resistor keeps the pin at 0V (LOW) when the button is not pressed. Pressing the button connects the pin to 3.3V (HIGH), so GPIO.input returns 1.

🚀 Application

expert

Counting button presses with GPIO.input

You want to count how many times a button connected to GPIO pin 18 is pressed. Which code snippet correctly increments count only when the button is pressed (assuming pull-up resistor)?

A

if not GPIO.input(18):
    count += 1

B

if GPIO.input(18) == 0:
    count += 1

C

if GPIO.input(18):
    count += 1

D

if GPIO.input(18) == 1:
    count += 1

💡 Hint

Remember that with a pull-up resistor, the pin reads 1 when not pressed and 0 when pressed.

With a pull-up resistor, the input reads 0 when the button is pressed. So if not GPIO.input(18): detects the press correctly.