simple question

Projects Programming
1

hey guys

I have the arduino being activated in a certain pin "trigger"
and then it activates a pin if "trigger" is HIGH
I'll put here a part of the code:

void loop(){

digitalWrite(x, HIGH); //set pin_number X to HIGH
delay(time_delay);
digitalWrite(x, LOW); //set pin_number X to LOW
delay(time_delay);
}

My question: when I run the arduino, it outputs several times to HIGH in "pin_number X" because it is on "void loop" function
but I want it to output only once, and then put that pin to LOW forever (until I reset it again)
I tried to put that code on the "void setup", but like that it doesn't output correctly (or doesn't even output anything)
Thank you

2

I would add a counter variable. Something like this (completely untested). The variable count will only eqaul zero once.

void setup(){
int count =0;
}

void loop(){
if (count==0)
{
digitalWrite(x, HIGH);  //set pin_number X to HIGH
    delay(time_delay);
count=1;
}
    digitalWrite(x, LOW);   //set pin_number X to LOW
    delay(time_delay);
}
3

I tried to put that code on the "void setup", but like that it doesn't output correctly (or doesn't even output anything)

That's the place I'd do it.

Of course, if we could see the code, we'd be able to help.

4

@Pauly:

I would add a counter variable. Something like this (completely untested).

Good thing you stated that. Creating the variable in setup() and trying to access it in loop() is doomed to failure. As a global, it would work. As a static variable, created in loop(), it would work. As written, it will not.