Question

When a byte from the serial port is received, it enters this handler correctly, but my label on the GUI does not change. Edit: Yes, it is in the same class as the GUI
Edit 2 The function declaration does not need to be ‘static’…I merely copy-pasted an example from msdn
edit 3 It works after getting rid of the static decleration and using something like this.
Thanks for your help

private static void DataReceivedHandler(object sender, SerialDataReceivedEventArgs e)
        {
            Form1 gui = new Form1(); //main GUI

            try
            {
                SerialPort sp = (SerialPort)sender;
                string indata = sp.ReadExisting();
                //gui.rxLabel.Text = indata;
                gui.txLabel.Text = "testttingg";
            }
            .......

Answer 1

Why are you declaring a new instance of your form? Just use your form’s txLabel

In c#:

this.txLabel.Text = "testing";

In vb.net:

Me.txLabel.Text = "testing"

In the sample code you posted you are creating a new instance / reference to your actual form. Use the existing instance, in addition, use this rather then a new instance.

I had to edit my question as I noticed you were using a static method.

Try this:

 public static void DataReceivedHandler(object sender, SerialDataReceivedEventArgs e)
        {
            Form1 f = (Form1) sender;

            f.textBox1.Text = "testing";

        }

        private void button1_Click(object sender, EventArgs e)
        {
            DataReceivedHandler(this, null);     
        }

Of course I am just calling DataReceivedHandler() from a button event, you can call it from your own event. Point being is you need to pass the current instance, this to the function. Once inside the function do not create a new instance of the form, just set a reference to the current form and use that reference to apply the settings to the property (txtBox or Label or whatever).