I am trying to get the Raspberry Pi to shut down when a GPIO input is triggered. I have a c version of a small program and a python version which both do what I want to do but both cause high CPU load.

The c example:

#include <wiringPi.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int main (void)
    wiringPiSetup ();
    pinMode (5, INPUT);
        if(digitalRead (5) == HIGH)
                printf("Sleepy Pi requesting shutdown on pin 24\n");
                system("sudo shutdown -h now");
     return 0;

Here the example in python:


import RPi.GPIO as GPIO
import os, time

GPIO.setup(24, GPIO.IN)
GPIO.setup(25, GPIO.OUT)
GPIO.output(25, GPIO.HIGH)
print ("[Info] Telling Sleepy Pi we are running pin 25")

while True:
        if (GPIO.input(24)):
                print ("Sleepy Pi requesting shutdown on pin 24")
                os.system("sudo shutdown -h now")

Is this the right approach to what I am trying to do or is there some other way getting this done without causing high CPU load ?

Kind Regards Jan

  • 1
    If you want to do your own; Rather than polling, create a process which waits for an interrupt, then run it as a background process. This will consume minimal resources.
    – Milliways
    Nov 19, 2017 at 10:45
  • There are many similar poweroff questions on this site. The process I use is raspberrypi.stackexchange.com/a/42945/8697, although if I were starting again I would explore the inbuilt gpio-shutdown which should do this (although I have not tried it).
    – Milliways
    Nov 19, 2017 at 11:53

1 Answer 1


Put a small delay in each while loop. Perhaps a sleep of a tenth or a hundredth of a second. That will test the GPIO 10 or 100 times a second but still give plenty of time for other activities.

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