Binaries

[Amon Ra]

Urgent. How do i convert a decimal to binaries and vice versa? Also, how can i fiund the high and low bit of the binary?

[mlewis]

Check out this thread:

http://forums.vb-world.net/showthrea...6&goto=newpost

The question is answered quite well there.

Function Bin(ByVal sBinary As String) As Long
    Dim I As Integer
    For I = 0 To Len(sBinary) - 1
    If Mid(sBinary, Len(sBinary) - I, 1) = "1" Then Bin = Bin + (2 ^ I)
    Next
End Function

Usage:

Print Bin("110001010101010001011011")

[Amon Ra]

in the following:

code:--------------------------------------------------------------------------------

1-Private Function Dec2Bin(Num As Integer) As String
2- Dim i As Integer
3- Dec2Bin = ""
4- For i = 0 To 7
5- If Num And 2 ^ i Then
6- Dec2Bin = "1" + Dec2Bin
7- Else
8- Dec2Bin = "0" + Dec2Bin
9- End If
10- Next i

End Function

what does the line 5 do?

[Amon Ra]

the function you, megatron, gave me does not work for me

if i convert 2 to binaries, it returns 00000010
but if i try to go from bin. to dec. with your function with
00000101, it returns 14 instead of 5 . i dont get it

2 is 00000010 in binary.

[mlewis]

OK, in decimal:

1234 is broken up like this: 1 thousand, 2 hundreds, 3 tens, and 4 ones. In binary, it works like this:

1111 is 1 one, 1 two, 1 four, and 1 eight, or 15, while 1110 is no ones, 1 two, 1 four, and 1 eight, or 14. Line 5 in that code checks to see whether the next digit it needs to write is a 0 or a 1.

[mlewis]

Sorry, Meggie, but I just had to say, do your homework.

Amon ra: the code should read as below:

Public Function Bin2Dec(Byval Bin As String) As Long
 Dim I As Integer, RetVal As Long
 For I=Len(Bin) To 1 Step -1
  If Mid$(Bin, I, 1)="1" Then RetVal=RetVal+(2^(I-1))
 Next I
End Function

This should work (quite) a bit better.

[Lord Orwell]

here is something i threw together a few min ago.

Function dec2bin(numtoconvert As Long) As String
   Dim worktext As String
   Dim cl As Long
   Dim workval As Long
   worktext = Oct(numtoconvert)
   For cl = 1 To Len(worktext)
       workval = Val(Mid(worktext, cl, 1))
      If (workval And 4) = 4 Then output = output + "1" Else output = output + "0"
      If (workval And 2) = 2 Then output = output + "1" Else output = output + "0"
      If (workval And 1) = 1 Then output = output + "1" Else output = output + "0"
   Next cl
   dec2bin = output
End Function

Function bin2dec(text As String) As Long
   Dim multiplier As Double
   Dim cl As Long
   multiplier = 1
   Dim output As Double
   For cl = Len(text) To 1 Step -1
      output = output + (Val(Mid$(text, cl, 1)) * multiplier)
      multiplier = multiplier + multiplier
   Next cl
   bin2dec = output
End Function

I hope you don't have any problem decoding HOW they work. Just be happy that they do work

[Amon Ra]

Thanks guys. Now i have to decide which one to use

[Lord Orwell]

you have to use mine!
Im the only one who gave a converstion TO binary.

converting FROM is the easy part...

Originally posted by mlewis
Sorry, Meggie, but I just had to say, do your homework.

I beg to differ, but my function does work.

Try this (in a Module)

Function Bin(ByVal sBinary As String) As Long
    Dim I As Integer
    For I = 0 To Len(sBinary) - 1
    If Mid(sBinary, Len(sBinary) - I, 1) = "1" Then Bin = Bin + (2 ^ I)
    Next
End Function

Now place this line in a CommandButton and it outputs 5 (not 14).

Print Bin("00000101")

[Amon Ra]

OK, your functions all work Thx