Confused about using NOT and AND operators

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:

Code:

---

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:

Code:

---

  '
  '
 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

So what is the confusion?

Code:

---

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

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

Perhaps what you wanted was

Code:

---

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.

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.

Think links like this wikipedia one might be helpful

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

Code:

---

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:(.

Quote:

---

Originally Posted by DataMiser

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

Perhaps what you wanted was

Code:

---

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?

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.