_DebugOut() changes @error when _DebugSetup() is commented out

[AndyS01]

There is a problem with the way _DebugOut() handles @error when the call to _DebugSetup() is commented out.

The problem happens when the code calls some function that sets @error to some zero or non-zero value, and immediatly after making the call, calls _DebugOut() to log the returned value of @error, and after that, checks @error.

With _DebugSetup() commented out, @error is always set to 3!

The workaround is to preserve @error in a variable, call _DebugOut(), then check the preserved variable instead of checking @error.

Example code before:

CopyItem($ar_from, $ar_to, $selndx_from, $anchor_to)
  _DebugOut("+++: CopyItem returned with @error = " & @error)
  If (@error == 0) Then
   Return (1)
  EndIf

Example code with the work-around:

CopyItem($ar_from, $ar_to, $selndx_from, $anchor_to)
        $err = @error
  _DebugOut("+++: CopyItem returned with @error = " & @error)
  If ($err == 0) Then
   Return (1)
  EndIf

Test code showing the problem:

#include <Debug.au3>
;_DebugSetup(@ScriptName & "_debug.txt", False)
test()
Func test()
Local $er1, $er2
test_0_func()
    $er1 = @error
_DebugOut("+++: test_0_func() returns: @error = " & @error)
    $er2 = @error
if (@error == 0) Then
        msgbox(0, "", "$er1="&$er1&",$er2="&$er2& ": test_0_func() returns 0")
    Else
        msgbox(0, "", "$er1="&$er1&",$er2="&$er2& ": test_0_func() returns non-zero")
    EndIf
test_1_func()
    $er1 = @error
_DebugOut("+++: test_1_func() returns: @error = " & @error)
    $er2 = @error
if (@error == 0) Then
        msgbox(0, "", "$er1="&$er1&",$er2="&$er2& ": test_1_func() returns 0")
    Else
        msgbox(0, "", "$er1="&$er1&",$er2="&$er2& ": test_1_func() returns non-zero")
    EndIf
EndFunc
Func test_0_func()
Return SetError(0, 0, 0)
EndFunc
Func test_1_func()
Return SetError(2, 2, 2)
EndFunc

[BrewManNH]

From the help file for _DebugOut

_DebugSetup() must be called before the first call of _DebugOut(), or it will return with @Error = 3.

[AndyS01]

I thought that the way to compile a script without _DebugOut() calls was to comment out the call to _DebugSetup().

If I have a local variable that I set, but only look at with a _DebugOut() call, when I comment out _DebugSetup(), I get a "variable declared but not used" warning when I run the script. That shows that the _DebugOut() calls are somehow commented out.

For example:

func abc()
local $xx = Time()
_debugout("+++: time is: " & $xx)
endfunc

Am I wrong?

Andy

Edited May 21, 2012 by AndyS01

[BrewManNH]

_DebugOut calls will still be called, they just won't do anything other than return the @error = 3 because you have commented out the _DebugSetup. There's no way around it, either find and delete all lines that use _DebugOut, or do as you've done and preserve the @error in a variable when you're compiling.