Hi,
Try this code:

Private Function SeparateString(str As String) As String
Dim x As Integer
Dim retStr As String

retStr = ""
For x = 1 To Len(str)
retStr =...

I see. I'll change it. Thanks.
If you think that your problem is solved, please, mark as solved this thread.

Regards

HI,
I see your point. But what he need is to separate some bits from a word. To do this, the easy way is to use masks. The mask thing is not an invention of mine.

Take a look here:...

And it is nice too, but as he say, he don't have string binary values. All that I know is the existence of two bytes. I've coded taking account of these.
As you said... Have fun ;)

Hi,
Well, I guess you have to shift right the variable fourBits 12 positions.
Try this (I am not really sure it will work, because I have not VB here)

Dim byteA As Long
Dim byteB As...

Hi,
Try this

Dim byteA As Long
Dim byteB As Long
Dim sumBytes As Long
Dim fourBits As Long
Dim twelveBits As Long

byteA = 32 ' 00100000b = 32d

Hi,
Is your problem solved?

Hi,
Think of a property just as if it were a regular variable. You can get or assign only one value at once.
If you need to pass more than one parameter to a property, instead use a regular Sub...

Hi,
First, this is not VB6 code. It is VB.Net.
To load an image, you must code:
Dim coinImage As Image
Dim imagePath As String = ""

If choice = 0 Then
...

Veena, you must be joking! :lol:
Don't you? :-|

Try this with your code "Hello world" and please post how many words your program find.

CountWords is the function coded inside the module "mdlWordCount.bas".

Did you open de prjWordCount.vbp file to load the project in the VB6 IDE or just the frm file?

Hi! Tell me why my program doesnt work so I can fix it.

I have uploaded a sample VB6 project to demonstrate the use of this function.

I hope it could help.
Private Function CountWords(strText As String) As Long
Dim x As Long
Dim words As Long
Dim lenStr As Long
Dim lenPrevWord As Long
Dim currentAscii As...