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
> Does anyone know why I get an error message when I use the following code
> (it
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>
> JimmyX
Stephany Young - 13 Oct 2007 07:47 GMT
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.
> Hi Jimmy,
>
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>>
>> JimmyX
Scott M. - 13 Oct 2007 14:39 GMT
I'm pretty sure that's what I said:
"your conversion problem stems from the compound statement"
"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."
-Scott
> Close but no cigar.
>
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>>>
>>> JimmyX
Jan Hyde (VB MVP) - 15 Oct 2007 11:30 GMT
"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
Scott M. - 15 Oct 2007 13:16 GMT
Doh!
> "Scott M." <s-mar@nospam.nospam>'s wild thoughts were
> released on Fri, 12 Oct 2007 23:36:34 -0400 bearing the
[quoted text clipped - 13 lines]
>
> https://mvp.support.microsoft.com/profile/Jan.Hyde