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Question by Killer2gudPL · Dec 21, 2013 at 08:30 AM · error

unity error error CS1525: Unexpected symbol `(', expecting `identifier'

Hi, as you can see I got a problem. Here's the code:

using UnityEngine; using System.Collections;

public class GUITest : MonoBehaviour {

 void OnGUI () {
     if (GUI.Button (new Rect (25, 25, 100, 30), "Button")) 
     {        
         Application.LoadLevel.(1);
     }
 }
 

}

Thanks for helping. (I am new in Unity)

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Answer by Benproductions1 · Dec 21, 2013 at 08:42 AM

Hello,

Firstly you should format your code when posting questions on Unity Answers. If you don't know how to do that, there's a link to the Unity Answers tutorial video on the right of this webpage.

In C# a . as a piece of code is compiled as one of two things:
Either as a literal for float's or double's such as

 6.0 //double precision floating point
 8.83f //floating point

Or as a separator for sub-"fields" of certain data structures such as classes

 Application.LoadLevel //The static LoadLevel function of the static Application class
 GUI.Button //The static Button function of the static GUI class


When you used it in your script like so:

 LoadLevel.(1);

The C# compiler thinks that you are trying to access a sub-"field" of the LoadLevel function, but you follow that with a (. Since ( is not a valid name character, the C# compiler throws a Compiler Error with "Unexpected Symbol".

To call functions in C# (and in most other languages) you use a () pair, directly after the function reference, containing the arguments for the function like so:

 function(arguments)

You must use the correct syntax when writing code or it will not work.

I suggest you do some (a whole lot) of reading about C# syntax and language features so that you understand how to use the language before you post 50 questions about syntax errors.

Hope this helps,
Benproductions1

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avatar image Killer2gudPL · Dec 21, 2013 at 09:42 AM 0
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Thank you! :)

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Answer by KellyThomas · Dec 21, 2013 at 08:39 AM

 Application.LoadLevel.(1);

Should be:

 Application.LoadLevel(1);
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