"Stephany Young" <noone@localhost> wrote in message
news:ehdo0TWDIHA.5856@TK2MSFTNGP04.phx.gbl...
> Close but no cigar.
>
> The result of exponention {^} is always a Double.
>
> Prior to a bitwise And operation, one of the operands MAY be converted to
> another type.
> If one of the operands is an Integer and the other is Decimal, Single,
> Double or String then that one is converted to a Long and the result of
> the bitwise And operation will also be a Long
>
> The documentation provides a definitive table for this rule for any
> combination of operand types.
>
> The big clue is that an 'implicit conversion' is not allowed so therfore
> one needs to do an explicit conversion.
>
> Dim J as integer
>
> If J And Convert.ToInt64(2^0) Then ... etc.
>
>
>
> "Scott M." <s-mar@nospam.nospam> wrote in message
> news:ODmnaoUDIHA.3980@TK2MSFTNGP03.phx.gbl...
>> Hi Jimmy,
>>
>> Before I get to your conversion code, I must just say that you've got to
>> approach VB .NET 2005 with the understanding that it is not just a new
>> version of VB 6. It is a completely different beast, it runs on an
>> entirely new runtime and although some of its' syntax is similar to VB 6,
>> there really isn't much it has in common with VB 6. That's why you'll
>> find lots of things that worked in VB 6 that don't in VB .NET.
>>
>> Also, you should turn Option Strict on and leave it on forever in VB
>> .NET. This setting makes VB .NET a type-safe language like C# and
>> although it will mean that you will have to write more code at times, it
>> will ensure that the code you have is much more bullet proof come
>> runtime.
>>
>> Now, addressing your If...Then logic:
>>
>> First, you should lose the "goto" as this was never considered good
>> programming practice (unless you go back to QBASIC). Place the code you
>> want to run if your condition is true in the true section of the If
>> construct or create a separate sub/function and simply call it from the
>> true portion of the If construct.
>>
>> Next, your conversion problem stems from the compound statement. Your If
>> statement, it translates to:
>>
>> If j and 2 then
>>
>> Which doesn't make much sense and the compiler sees two different data
>> types trying to be compared. Since one side of your compound statement
>> will return a Double and the other side will return a Long and the
>> compiler won't do impicit conversions for you, you must make sure both
>> sides of the compound statement have the same type. This wouldn't be
>> necessary if both sides were expressions (if j <> 0 And 2^0 = 2 then your
>> code would work just fine)
>>
>> If you can explain what it is you are realy trying to test, I can help
>> you write the If logic more clearly.
>>
>> -Scott
>>
>>
>> "JimmyX via DotNetMonster.com" <u37963@uwe> wrote in message
>> news:799a5ec0c1856@uwe...
>>> Does anyone know why I get an error message when I use the following
>>> code (it
>>> works with VB6):
>>>
>>>
>>> Dim J as integer
>>> If J And (2^0) then goto ........
>>>
>>> I get an message that implicit conversions from Double to Long are not
>>> allowed. What is the proper
>>> VB2005 way to implement this statement. I don't want to have to keep
>>> Option
>>> Strict Off in order for
>>> my program to work!
>>>
>>> JimmyX
>>>
>>> --
>>> Message posted via http://www.dotnetmonster.com
>>>
>>
>>
>

>
>Next, your conversion problem stems from the compound statement. Your If
>statement, it translates to:
>
>If j and 2 then

"Jan Hyde (VB MVP)" <StellaDrinker@REMOVE.ME.uboot.com> wrote in message
news:u7g6h3pnnne6amu83ko84ltpok25u0tf7d@4ax.com...
> "Scott M." <s-mar@nospam.nospam>'s wild thoughts were
> released on Fri, 12 Oct 2007 23:36:34 -0400 bearing the
> following fruit:
>
> <SNIP>
>>
>>Next, your conversion problem stems from the compound statement. Your If
>>statement, it translates to:
>>
>>If j and 2 then
>
> If j and 1 Then
>
>
>
> --
> Jan Hyde
>
> https://mvp.support.microsoft.com/profile/Jan.Hyde