LCD display code problem

rascupanamuha · Jul 14, 2013

I am stuck on one simple thing.

I have 20x04 LCD display with IC2 board (the last one here): **broken link removed**

Here is what i want. If digital pin 6 is 5V, then write 5 on LCD, or if pin 6 is 0V, then write 0

void loop()
{
lcd.home ();
lcd.clear();
int x

;
if (LED2==HIGH) x=5;
else x=0;

lcd.setCursor ( 0, 2 );
lcd.print (x);

delay (200);
}

LCD always shows 0, even when it is not. And even when i reset atmega 328p

Please help me if you know what is wrong in my code

strykeroz · Jul 14, 2013

Hi,

Can you please post the rest of your code? What sets D6 high or low?

Geoff

rascupanamuha · Jul 15, 2013

Here is the whole code:

#include <Wire.h>
#include <LiquidCrystal_I2C.h>
int REED = 4;
int LED = 8;
int LED2 = 6;
int x = 9;

LiquidCrystal_I2C lcd(0x20, 2, 1, 0, 4, 5, 6, 7, 3, POSITIVE); // Set the LCD I2C address

void setup()
{
Serial.begin(9600); // Used to type in characters
lcd.begin(20,4); // initialize the lcd for 20 chars 4 lines and turn on backlight
lcd.backlight(); // finish with backlight on

lcd.setCursor(3,0); //Start at character 4 on line 0
lcd.print("Hello");
}

void loop()
{
lcd.home ();
lcd.clear();
int x;

if (LED2==HIGH) x=5;
else x=0;

lcd.setCursor ( 0, 2 );
lcd.print (x);

delay (200);
}

What sets D6 high or low? ME I manually put DIGITAL pin 6 (PIN 12 fisically) to GND or to +5V..
It is normally 0V, and it display it says 0, but when i put 5V to it, it still says 0..

Ian Rogers · Jul 15, 2013

Code:

int LED2 = 6;

LED2 will ALWAYS be high... You assigned the variable a 6...

You need to assign a pin to an output..

Code:

pinMode(LED2, OUTPUT);

Then control it... Set it high..

Code:

digitalWrite(LED2, HIGH);

Or low..

Code:

digitalWrite(lLED2, LOW);

I don't know whether you can read a pin whilst in output mode..

ledState = digitalRead(LED2);

If not you'll need an output buffer so you can read the buffer instead of the pin...

strykeroz · Jul 15, 2013

Hi,

Ian's right on the money. Testing the value 6 will always equate to high.

To get to where you want to be with reading a pin you're externally toggling,

Code:

if(digitalRead(LED2)==HIGH) x=5; else x=0;

will do the trick. Or, since HIGH and true are the same internally to an Arduino, you could simplify this to

Code:

if(digitalRead(LED2)) x= 5; else x = 0;

You can read a pin in output mode so that last part is fine. I'll often use

Code:

digitalWrite(!digitalRead(LEDPIN));

to toggle a LED from whatever state is it is to the alternative on an output pin and that's all good.

Cheers !
Geoff

rascupanamuha · Jul 15, 2013

You are awesome ! Thank you

Here is what i did, and now it works great. Finally!

#include <Wire.h>
#include <LiquidCrystal_I2C.h>
int REED = 4;
int LED = 8;
int LED2 = 6;
int x = 9;

LiquidCrystal_I2C lcd(0x20, 2, 1, 0, 4, 5, 6, 7, 3, POSITIVE); // Set the LCD I2C address

void setup()
{
pinMode(LED2, INPUT);
Serial.begin(9600); // Used to type in characters
lcd.begin(20,4); // initialize the lcd for 20 chars 4 lines and turn on backlight
lcd.backlight(); // finish with backlight on

lcd.setCursor(3,0); //Start at character 4 on line 0
lcd.print("Hello");
delay(500);
}

void loop()
{
lcd.home ();
lcd.clear();
int ledState;
int x;
ledState = digitalRead(LED2);

if (ledState==HIGH) x=5;
else x=0;

lcd.setCursor ( 0, 2 );
lcd.print (x);

delay (200);
}

That pin 6 is input pin, it will be connected to the switch, so i dont need digitalWrite on it but i get the point

LCD display code problem

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