IsInt() Freaking Out?

[=sinister=]

Func ConvertVariable($Variable)
    Local $Return
    Local $String = StringSplit($Variable, "=")
    Local $NewString1 = StringReplace($String[2], '"', "")
    Local $NewString = StringReplace($NewString1, "'", "")
    Local $Test = 1
    If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")
    If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")
    If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")
    
    If IsInt($NewString) Then
        MsgBox(0, "Int", $NewString)
        $Return = "int " & $String[1] & "=" & $NewString & ";"
    ElseIf StringInStr($NewString, "true") or StringInStr($NewString, "false") Then;IsBool() didn't seem to work?? =\
    If $NewString = "true" Then
        $Return = "bool " & $String[1] & "=" & "true" & ";"
        Else
        $Return = "bool " & $String[1] & "=" & "false" & ";"
    EndIf
;ElseIf IsString($NewString) Then
;       $Return = "char " & $String[1] & "=" & $NewString & ";"
    EndIf
    Return $Return
EndFunc

This is very weird to me.

Take a look at this code:

Local $Test = 1

If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")

If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")

If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")

$Test = 1. We confirm this is a int. IsInt($Test) works because $Test = 1. Then we check if $NewString = 1. It is, and I get the messagebox. However, IsInt($NewString) won't do anything? I'm 100000% confused.

Also, the parameter for the function is a variable without the $, I have this function used with some other code that trims the $ out.

[Malkey]

Func ConvertVariable($Variable)
    Local $Return
    Local $String = StringSplit($Variable, "=")
    Local $NewString1 = StringReplace($String[2], '"', "")
    Local $NewString = StringReplace($NewString1, "'", "")
    Local $Test = 1
    If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")
    If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")
    If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")
    
    If IsInt($NewString) Then
        MsgBox(0, "Int", $NewString)
        $Return = "int " & $String[1] & "=" & $NewString & ";"
    ElseIf StringInStr($NewString, "true") or StringInStr($NewString, "false") Then;IsBool() didn't seem to work?? =\
    If $NewString = "true" Then
        $Return = "bool " & $String[1] & "=" & "true" & ";"
        Else
        $Return = "bool " & $String[1] & "=" & "false" & ";"
    EndIf
;ElseIf IsString($NewString) Then
;       $Return = "char " & $String[1] & "=" & $NewString & ";"
    EndIf
    Return $Return
EndFunc

This is very weird to me.

Take a look at this code:

Local $Test = 1

If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")

If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")

If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")

$Test = 1. We confirm this is a int. IsInt($Test) works because $Test = 1. Then we check if $NewString = 1. It is, and I get the messagebox. However, IsInt($NewString) won't do anything? I'm 100000% confused.

Also, the parameter for the function is a variable without the $, I have this function used with some other code that trims the $ out.

If you add these three If statements, it should show what is happening.

If IsString($NewString) Then MsgBox(0, "This", "IsString Is  WORKING")
If IsInt(StringStripWS($NewString,8)) Then MsgBox(0, "This", "IsInt Is Not WORKING")
If StringIsInt(StringStripWS($NewString,8)) Then MsgBox(0, "This", "StringIsInt Is WORKING")

[martin]

Func ConvertVariable($Variable)
    Local $Return
    Local $String = StringSplit($Variable, "=")
    Local $NewString1 = StringReplace($String[2], '"', "")
    Local $NewString = StringReplace($NewString1, "'", "")
    Local $Test = 1
    If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")
    If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")
    If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")
    
    If IsInt($NewString) Then
        MsgBox(0, "Int", $NewString)
        $Return = "int " & $String[1] & "=" & $NewString & ";"
    ElseIf StringInStr($NewString, "true") or StringInStr($NewString, "false") Then;IsBool() didn't seem to work?? =\
    If $NewString = "true" Then
        $Return = "bool " & $String[1] & "=" & "true" & ";"
        Else
        $Return = "bool " & $String[1] & "=" & "false" & ";"
    EndIf
;ElseIf IsString($NewString) Then
;       $Return = "char " & $String[1] & "=" & $NewString & ";"
    EndIf
    Return $Return
EndFunc

This is very weird to me.

Take a look at this code:

Local $Test = 1

If IsInt($Test) Then MsgBox(0, "Test", "Test is = 1... and IsInt() is working")

If $NewString = 1 Then MsgBox(0, "TEST", "ok, it IS a freaking Int, why isn't IsInt() working?!")

If IsInt($NewString) Then MsgBox(0, "This", "Is Not WORKING")

$Test = 1. We confirm this is a int. IsInt($Test) works because $Test = 1. Then we check if $NewString = 1. It is, and I get the messagebox. However, IsInt($NewString) won't do anything? I'm 100000% confused.

Also, the parameter for the function is a variable without the $, I have this function used with some other code that trims the $ out.

The problem is that AutoIt is too smart.

You have $Test as an integer.

You have $NewString as a string. So it will not be an inetegr using IsInt because it's a string type.

When you try to compare a string to an int then AutoIt will try to convert the string to an integer while it makes the comparison so you can say

If "7" = 7 then ConSoleWrite("AutoIt is too smart" & @CR)

but "7" is still a string and 7 is still an integer.

[jvanegmond]

To do any type of If checking in AutoIt, I can recommend using == instead of a single =.

The == logic is very straightforward, and easy to understand.