X=1 .... If X=0 and Z() Then

[andyswarbs]

Can I ask fora point of clarification on the following:

Version 1:

$X=1
IF $X=0 and Z() Then
   Exit
Endif

My question "Is the function Z ever evaluated?". This is important since Z may be time-consuming. So a more laborious bit of code but safer, might be:

Version 2:

$X=1
If $X=0 Then
  If Z() Then
    Exit
  Endif
endif

But is the latter necessary? And is Version 1 a happenstance, or (my ideal answer) intentional in the language. I how safe am I in using version 1.

Regards

Andy

Edited March 3, 2006 by andyswarbs

[greenmachine]

Here are 3 tests I tried. It appears that Z() is never executed if $x does not pass the test.

$X=1
IF $X=0 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 0
EndFunc

MsgBox (0, "exiting", "normally")

[Nuffilein805]

Here are 3 tests I tried. It appears that Z() is never executed if $x does not pass the test.

$X=1
IF $X=0 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    Sleep (5000)
    Return 0
EndFunc

MsgBox (0, "exiting", "normally")

there is a difference in z and Z

if you try it with z() instead of Z() you will get inside the function

$x = 0
if $x = 1 and z() then
...
endif

won't get inside the function

so its up to you to choose how you write your code

[SmOke_N]

there is a difference in z and Z

if you try it with z() instead of Z() you will get inside the function

$x = 0
if $x = 1 and z() then
...
endif

won't get inside the function

so its up to you to choose how you write your code

I'm not sure I understand what you mean a difference between Z() and z() in this post:

MsgBox(0, '', Z())
Func z()
    Return 5
EndFunc

MsgBox(0, '', y())
Func Y()
    Return 6
EndFunc

Edit:

Or even trying to reitterate that it does reach Z()... or z() ... $x obviously = 0 but Z() obviously would return 5, so if one of the 2 are true then it should give you the msgbox:

$x = 0
If $x = 1 Or Z() = 5 Then MsgBox(0, 'Done', 'Reached Z() and it = ' & Z())
Func z()
    Return 5
EndFunc

Edited March 3, 2006 by SmOke_N

[Jon]

Can I ask fora point of clarification on the following:

Version 1:

$X=1
IF $X=0 and Z() Then
   Exit
Endif

My question "Is the function Z ever evaluated?".

No Z() isn't evaluated.

[greenmachine]

there is a difference in z and Z

if you try it with z() instead of Z() you will get inside the function

$x = 0
if $x = 1 and z() then
...
endif

won't get inside the function

so its up to you to choose how you write your code

Did you try it? It doesn't matter if it's small z or big z.

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    MsgBox (0, "in z", "z = Z")
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

I definitely got in z() there.

Plus, if you use z() and Z() (which according to you would be different), you get this error:

ERROR: Z() already defined.

Func Z()

~~~~~~~~^

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    MsgBox (0, "in z", "z = Z")
    Return 1
EndFunc

Func Z()
    MsgBox (0, "in big Z", "wtf")
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

This shows that AutoIt considers z() and Z() to be the same function, and this makes perfect sense since it's not case-sensitive.

[SmOke_N]

Did you try it? It doesn't matter if it's small z or big z.

$X=1
IF $X=1 and Z() Then
    MsgBox (0, "in func", "whee")
   Exit
Endif

Func z()
    MsgBox (0, "in z", "z = Z")
    Return 1
EndFunc

MsgBox (0, "exiting", "normally")

I definitely got in z() there.

Plus, if you use z() and Z() (which according to you would be different), you get this error:

$X=0 ; this one will not go to z() because $x = 0

IF $X=1 and Z() Then

MsgBox (0, "in func", "whee")

Exit

Endif

Your example shows $x = 1 as true which it is, so then it goes to Z(). Now if $x = 0 then it is false, so it never goes to Z() because the first inquiry isn't true so there is no need to go to z().

[greenmachine]

A quote from myself when I first posted: Z() is never executed if $x does not pass the test.

What I was showing in my last post is that when $x passes (and the script goes on to check z()), it doesn't matter if the functions match case or not. z() = Z(). That's all I'm saying.

I made $x = 1 on purpose the second time around, so that I knew it would be going to Z(). I was hoping everyone would understand that the different tests were done on purpose....

[SmOke_N]

A quote from myself when I first posted: Z() is never executed if $x does not pass the test.

What I was showing in my last post is that when $x passes (and the script goes on to check z()), it doesn't matter if the functions match case or not. z() = Z(). That's all I'm saying.

I made $x = 1 on purpose the second time around, so that I knew it would be going to Z(). I was hoping everyone would understand that the different tests were done on purpose....

I understood, I was reitterating your post .