So I have just an led and an Resistor connected to Pin 2 and my code is this:
void setup() {
pinMode(2, OUTPUT);
}
void loop() {
digitalWrite(2, HIGH);
delay(500);
digitalWrite(2, LOW);
}
However the Pin just stays in either HIGH or LOW depending on which came first in the code (I switched it around a few times). But this code works, so I am wondering, why just why
digitalWrite(2, !digitalRead(2));
delay(500);
Try:
void loop() {
digitalWrite(2, HIGH);
delay(500);
digitalWrite(2, LOW);
delay(500);
}
1 Like
You don't have a delay after you set the pin the second time. It will briefly go to that state and then back to the original state.
1 Like
oh my god I am so stupid, thank you very much for answering my dumb question...
system
Closed
November 22, 2023, 9:23pm
5
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