The code of ISR executes twice

Question

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?

• JRobert: 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?
• explorer: @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..
• Gil: 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.

Answer 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.

• Paolo Zanchi: Remember to mark your answer as solution in three days ;)