Serial.OnGUI C# (CSharp) Метод

    void OnGUI()
    {
        // Show debug only if enabled and by the first instance to avoid overwrite same data
        if (s_debug && this == s_instances[0]) {
            GUILayout.Label("Serial last data: " + s_lastDataIn + " (last check: " + s_lastDataCheck + ")");
        }
    }