Confused about using NOT and AND operators

[jmsrickland]

I have two If statements using the NOT operator:

1) If Not 3 And &H88 Then

2) If Not 15 And &H88 Then

In both cases control goes into the If clause.

What confuses me is the result of (1) is 0 (3 And &H88 = 0) which to me would cause control to enter the If clause but the results of (2) is 8 (15 And &H88 = 8) which to me should skip the If clause and exit out.

I'm trying to mimic some C code and the C code is this:

 if(!((isqTo = isqFrom + *pvec) & 0x88))
 {
    '
    '
 }

Now when the above is executed the first time isqFrom = 4, *pvec points to memory that contains the value -1 and the results in isqTo is 3 and after the & 0x88 control enters the if clause just like it does in the VB example. Now the second time the if is executed isqFrom = 1, *pvec = 14 and isqTo = 15 and after the & 0x88 control skips out of the if statement.

So, in the VB I have this:

  '
  '
 isqTo = isqFrom + pvec(i)

 If isqTo > 0 Then
   '
   ' 1st time isqTo is 3
   ' 2nd time isqTo is 15 
   If Not isqTo And &H88 Then
       '
       ' 
       '
   End If
 End If
    '
    '

The variables in the VB code are the same as they are in the C code

[DataMiser]

So what is the confusion?

Debug.Print 3 And &H88
Debug.Print 15 And &H88

Yields

0
8

Not 0 would be true
Not 8 would be false

As 0 is false and anything other than 0 is true the Not gives the opposite result

[DataMiser]

After a couple of tests I think I see where the confusion is.

Perhaps what you wanted was

If Not CBool(isqTo And &H88) Then
       '
       ' 
       '
   End If

I really would have thought the code you had would have did this but after testing it I did not see the result I expected.

[passel]

Since all the values are numeric, the code is doing a binary Not, so is flipping all the bits, so the result of (15 And &H88) is non-zero and the inverse of it is still non-zero.
Changing the non-zero value to a boolean first, then does a logical Not, so non-zero inverted become logically a 0, i.e false.

Its another example of C being clear with two operators, and VB being context sensitive because it uses the same operator for two different operators.
In C you boolean logical Not "!", and binary operator Not "~".
In VB you use "And" for both, and the context of the variables operated on determine whether it is logical Not or binary operation Not.

[LaVolpe]

Think links like this wikipedia one might be helpful

[Jonney]

FYI, I just made a mistake by using Not Len(text):

Dim text As String
   text = "A"
   If Not Len(text) Then
      Debug.Print "Empty"
   Else
      Debug.Print "Has Text"
   End If

I 'expect' "Not Len(text)" is equal with "If Len(text)=0 then", but VB Slapping me.

[Nightwalker83]

After a couple of tests I think I see where the confusion is.

Perhaps what you wanted was

If Not CBool(isqTo And &H88) Then
       '
       ' 
       '
   End If

I really would have thought the code you had would have did this but after testing it I did not see the result I expected.

Wouldn't the not only be applied to "isqTo" in that case?

[passel]

Are you talking about "Not CBool(isqTo And &H88)" ?
Both numbers are in parentheses so will be operated on first.
isqTo is a number, and &H88 is a number, so those will be Binary Anded together, and the result converted to a boolean (True or False) by the CBool function, and that boolean logically negated to (False or True) by the Not operator.
The two examples given,
(3 And &H88) result is 0, so 0 is converted to False (0), then the Not returns True (-1).
(15 And &H88) the result is 8. so 8 is converted to True (-1) and the Not returns False (0).

Without the CBool
(15 And &H88) gives 8, and the Not complements (ones) the bits, so becomes -9, and any non-zero value will be evaluated as "True" in an If statement, so both (3 and &H88) and (15 and &H88) would be evaluated as True by the If without the CBool.

If you're talking the original OP example,
If Not isqTo And &H88 Then

Then, yes the Not inverts isqTo first, and Ands it with &H88, giving a different set of numbers, but still a non-zero result in a lot of cases that will be evaluated as True by the If, such is the case of both examples of 3 and 15.
Of course, 3 is being inverted to -4, and Anded with &H88 resulting in 136, which is evaluated as True because it is non-zero, not because it was 0 (false) and logically inverted to True, so the assumption that 3 and &H88 was "working" wasn't really true.