Can A Variable Set Another Variable?

[z0mgItsJohn]

Ok... I Am Working Of A System And I Want The Func To Be Able To Set A Return...

Heres What I'm Talking About...

Func _Add ($Add_1 , $Add_2 , $Return)
$Return = $Add_1 + $Add_2
EndFunc

;Example

_Add (5 , 5 , $Add_Equal ) ; I Want $Add_Equal To Be The Return Value.... Without Haveing To Set It In The Top
MsgBox (0, "Test", $Add_Equal,0)

Can Anyone Help Me Please! :/

Please Give Me An Example.. If You Can Help Me...

Edited January 10, 2008 by John2006

[Xenobiologist]

HI,

Func _Add($Add_1, $Add_2)
    Return $Add_1 + $Add_1
EndFunc  ;==>_Add

;Example

; I Want $Add_Equal To Be The Return Value.... Without Haveing To Set It In The Top
MsgBox(0, "Test", _Add(5, 5), 0)

So long,

Mega

[Fossil Rock]

Func _Add ($Add_1 , $Add_2 , $Return)
$Return = $Add_1 + $Add_1
Return ($Return)
EndFuncoÝ÷ Ú«¨µéÚ

Can Anyone Help Me Please! :/

Please Give Me An Example.. If You Can Help Me...

[z0mgItsJohn]

That Dose Not Answer My Question! :/

[z0mgItsJohn]

Ahh! Fossil Rock You Posted Before I Was About To Post The One Above The One Above Is For Xenobiologist

[z0mgItsJohn]

Can I Make $Return = A Different Var... Without Edited The Func? Please Help >.<

[z0mgItsJohn]

Well I'm Off To Bed Ill Check The Topic Tomorrow @ School ^^

But Please Keep Giving Answers ^^

Edited January 10, 2008 by John2006

[DW1]

This must be a typo

$Return = $Add_1 + $Add_1

... $1 + $1? you mean $1 + $2 right?

If you are going to hardcode the "$Add_Equal" in

"_Add (5 , 5 , $Add_Equal )"

then why not just hardcode the return:

Func _Add ($Add_1 , $Add_2 )
Return $Add_1 + $Add_2
EndFunc
$Add_Equal = _Add (5 , 15 )
MsgBox (0, "Test", $Add_Equal,0)

[DirtDBaK]

@John2006: there is an Edit button.

But Anyways if you could tell what you mean by:

$Return = A Different Var... Without Edited The Func?

If you mean that You want to use the $Return variable outside of the function and inside of the function it would be something like this:

Global $Return
_Add(1,1) ;This should equal 2
MsgBox(0,'',$return) ;should be 2
$Return = $Return + 7
MsgBox(0,'',$return) ;2+7 = 9 therefore the answer should be nine
_Add(3,4); 3+4 = 7 so the answer would be 7
MsgBox(0,'',$return)




Func _Add($var_1,$var_2) ;Adds var 1 and Var 2 into the $return var
$Return = $var_1 + $var_2
Endfunc

Notice this way the var $return doesn't have to be just inside the fucntion, nor do you have to pass it. See the globals work so that functions read the var the same way the script outside of functions do. The code you posted has local vars thats why you must use Return() to return that local var. Hope that helps!

[rover]

Local $Add_Equal

_Add (5 , 5 , $Add_Equal)
MsgBox (0, "Test", $Add_Equal,0)

Func _Add ($Add_1 , $Add_2 , ByRef $Return)
    $Return = $Add_1 + $Add_2
EndFunc

Edited January 10, 2008 by rover

[z0mgItsJohn]

Local $Add_Equal

_Add (5 , 5 , $Add_Equal)
MsgBox (0, "Test", $Add_Equal,0)

Func _Add ($Add_1 , $Add_2 , ByRef $Return)
    $Return = $Add_1 + $Add_2
EndFunc

Thank You So Much Rover ^^ My Question Was Answered

The Topic Can Be Closed Now ~~