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Question by 3dDude · Feb 20, 2013 at 05:54 AM · javascriptfunctionclassesarguments

Function argument referencing variables?

Hello everyone,

I have been trying to optimize some of my code by merging similar sections.

I have a function which will make some changes to a certain variable.

 function changeVariable (test : someClass) {
     test.testThing = gameObject;
     test.othertestThing = 10;
     
 }

something like that, obviously this isn't the actually code.

I wan't to be able to call this function for different variables.

 class someClass{
     var testThing : GameObject;
     var othertestThing : float;
 }
 
 
 var testsomeClass : someClass;
 var testsomeOtherClass : someClass;
 
 changeVariable(testsomeClass);
 //or
 changeVariable(testsomeOtherClass);

Is there any way to do this without writing the changeVariable() code twice?

Cheers

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Answer by pakfront · Feb 20, 2013 at 06:46 AM

I wan't to be able to call this function for different variables.

Pardon me if I am missing something obvious or misunderstanding your question, but have you tried your example code? It should work the way you want as is. In fact, that is the whole point of functions... to be able to run the same code on different variables interchangeably.

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avatar image 3dDude · Feb 20, 2013 at 05:06 PM 0
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I have indeed tried it. Getting the variable to the function is not the problem... but rather writing it back to the original variable.

if I have an array for example.

 var someArray : someClass[];

and I say:

 changeVariable(someArray[1]);

 function changeVariable (test : someClass) {
     test.testThing = gameObject;
    //this will not change someArray [1] to equal this gameObject will it?
 }
avatar image Dave-Carlile · Feb 20, 2013 at 05:15 PM 0
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As long as someClass is a class, and not a struct, then yes, that should work properly. You're referencing properties of the class, so any changes you make will be fine. Now, if you were to change the test variable itself (e.g. test = someNewObject) then no, that change wouldn't make it back to the caller.

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