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i am using Arduino Mega.. ANd this is the code:

void bump(void);

void setup() 
{
    pinMode(19,INPUT);
    attachInterrupt(4,bump,HIGH);
    pinMode(13,OUTPUT);
}

void loop() 
{

}

void bump()
{
   for(float i = 0; i < 90000; i++){
   digitalWrite(13,HIGH);
   }
   for(float i = 0; i < 10000; i++){
   digitalWrite(13,LOW);
   }

}  

Although i give +5 V to it once, the LED will blink twice...! What is the solution?

3
  • Why have you chosen float as the data type of your counter, 'i'? At least, is 'float' used elsewhere in your program? If not, this use drags in a lot of code, just for counting. And why are you re-writing your port i/o on each count? Wouldn't: digitalWrite(13,HIGH); delay(ONTIME); digitalWrite(13,LOW); delay(OFF_TIME); do what you intend?
    – JRobert
    Commented Apr 24, 2015 at 20:03
  • @JRobert: I wanted a huge value for i so i chose float.. Mayb long would have been better.. And we can't use delay inside ISR..
    – explorer
    Commented Apr 25, 2015 at 6:02
  • ISRs should never do more then needed. You can block other functions of the Arduino. Very seldom do I do anything more then set or clear a flag. In some variations of the Arduino family long times will cause it to crash.
    – Gil
    Commented Sep 24, 2020 at 21:15

1 Answer 1

1

After wondering over it for quite a long time, I finally came to the conclusion, that the interrupt should be made to trigger at RISING edge, if I want it to go into ISR the moment it gets HIGH.

Hence the setup line was change to

attachInterrupt(4,bump,RISING);

And the cod e works well.
I suppose that is because, when it receives a HIGH signal, than the period for which it receives HIGH is a bit long, during which it adds another interrupt into the queue, and so the ISR is executed once again.

2
  • Remember to mark your answer as solution in three days ;) Commented Apr 24, 2015 at 7:48
  • Oh yes, thanks a lot.. I didn't knew that :)
    – explorer
    Commented Apr 24, 2015 at 7:49

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