system
January 31, 2013, 12:14am
1
Hi all....
int voltone = 2;
int volttwo = 4;
void setup() {
Serial.begin(9600);
pinMode(voltone, OUTPUT);
pinMode(volttwo, OUTPUT);
}
void loop() {
digitalWrite(voltone, HIGH);
delay(1000);
digitalWrite(volttwo, HIGH);
delay(1000);
digitalWrite(voltone, LOW);
digitalWrite(volttwo, LOW);
}
what am i expecting from this code is to set one "LED"(for example) on then wait then set the other "LED 2" on then turn off LED and finally turn off LED2 .... but that is not what i'm getting! ..... what i'm getting is this.... LED is on and remain on while LED2 blinks with no end...
please help
system
January 31, 2013, 12:17am
2
Think about what your program is doing:
forever:
Waiting a bit after turning them off might matter.
system
January 31, 2013, 12:27am
3
Your code is good! Up to a point.
What is happening is that after you turn both LEDs OFF, you need another delay to see it. Just add another delay(1000) at the bottom of the code.
What is happening right now is that LED1 is turned OFF at the bottom of the code and right back ON at the top.
Good luck!
system
January 31, 2013, 12:31am
4
ahaa.... so its just the delay which will make me see what actually happens
system
January 31, 2013, 12:48am
5
Yes. Another way to look at it is that the time between the end of "void loop()" and the beginning of it, is almost zero. That's the "loop" part
