Retrieving bits in VB 6

Evil Phil

Does anybody know of a way to retrieve bits in VB6 from an integer? Basically I want to retieve the first bit to see if its 1 or 0, retrieve the second bit to see if its 1 or 0, and so on for the full 16.

Also signed datatypes: both Integer and Long in Vb6. Does anybody know which bit is set to specify the sign (+/-)?

seamus

This is probably a bit overkill, I'm sure there's better ways of doing it, but in Java...

// int a is the integer you want to check

boolean [] vals = new boolean[16]

for(int i = 15; i >= 0; i--) {
    if (a >= (2^i)) {
         a -= 2^i;
         vals[i] = true;
     } else
          vals[i] = false;
}

I'm sure you can read that, but basically, starting at bit n, if the total of the number is >= the value of 2^n, then bit n is 1, so reduce the value of the number by 2^n and continue for all n down to 0. If total of number is less than 2^n, then bit n is 0, so you just record the 0 and continue.

At the end you should have an array full of true and false values, etc.

That's the way I'd do it

Evil Phil

Its nearly a couple of years since I worked with java last. In Java ^ is an Xor is it not?

bonkey

Originally posted by Evil Phil
Does anybody know of a way to retrieve bits in VB6 from an integer? Basically I want to retieve the first bit to see if its 1 or 0, retrieve the second bit to see if its 1 or 0, and so on for the full 16.

Also signed datatypes: both Integer and Long in Vb6. Does anybody know which bit is set to specify the sign (+/-)?

Have a look at this : http://www.experts-exchange.com/Programming/Programming_Languages/Visual_Basic/Q_20151207.html

Should give you most of what you want on retrieving/manipulating bits. There's a method in one of the final posts called "IsBitSet" which should give you what you want I think.

As for the structure....VB stores the sign in the left-most bit, and stores its numbers with the most significant digit first (i.e. the same way we write numbers)

jc

Talliesin

Originally posted by bonkey
As for the structure....VB stores the sign in the left-most bit, and stores its numbers with the most significant digit first (i.e. the same way we write numbers)

Kind of. Generally we would consider that to be the order, i.e. we would generally talk about/write about/draw the bits of a long as being written that way if so that the number would be:

11111111111110010110111000000101

The first bit in the list (normally labelled as bit 31) is the sign bit.

However the order in memory is a different matter. Breaking the above into octets we have:

11111111|11111001|01101110|00000101

This is the correct way to talk about the mathematics of what is going on, but isn't what's happening in the memory.

If you have to worry about memory representations (rare in VB, but not unheard of) then it isn't meaningful to talk about bit orders (you can't address them) so we might as well use hex for more readability:

FF|F9|6E|05.

The actual order in memory will be little-endian:

05|6E|F9|FF.

This becomes important if you are using long values in conjuction with CopyMemory, MoveMemory or kludgy struct-based conversions.

Evil Phil

If (a And (2 ^ i)) then
    Debug.Print 1
Else
    Debug.Print 0
End If

Theres sample code equivilant to what I ended up doing. Here a is the integer containing the bits and i is the bit being checked for.

What I'm doing is storing the values of a number of checkboxes in a single integer and then retrieving them. I could go into detail but not unless you really, really want me too.

But what I'm actually doing is developing an algorithm that has serveral uses.

Talliesin

The problem with that pseudo code is that there is no concept of an unsigned type in VB (except for bytes, for which there is no concept of a signed equivalent).

The following overcomes this for Longs

Public Function BitTestLong(a As Long, i As Long) As Boolean
    Const HighestBit As Long = 31
    Select Case i
        Case Is < 0, Is > HighestBit
            Err.Raise 5
        Case HighestBit
            BitTestLong = a < 0
        Case Else
            BitTestLong = a And (2 ^ i)
    End Select
End Function

For Integers you would change the type of a to Integer, i can be left long (since it's the most efficient type), you would also change HighestBit to have a value of 15.

Talliesin

It might be faster to have a lookup array of the powers of 2 (0, 1, 2, ..., &H40000000, &H80000000) and use that.