[RESOLVED!] Need an explanation for this (decimal -> binary)

Hi, i want to convert decimal numbers to binary numbers.
this code here is working, only I dont understand it really... can sb explain
what this code does exactally?

and 2de question, I just can't find a code wich converts binary to decimal :sick:
sb knows a code for it???

Code:

---

Dim Number As Integer
Dim Binary_String As String
Number = Val(Text1.Text)
Binary_String = ""
While Number > 0   
Binary_String = Str(Number Mod 2) & Binary_String   
Number = Number \ 2
Wend
Text2.Text = Binary_String

---

VB Code:

Dim Number As Integer
Dim Binary_String As String
 
Number = Val(Text1.Text)
Binary_String = ""
Do While Number > 0
    Binary_String = Number Mod 2 & Binary_String
    Number = Number \ 2
Loop
Text2.Text = Binary_String

This is taking an integer and creating a string of 0s and 1s for the binary representation of that number.

For example, decimal 13 is binary 1101.
Within the loop, see if the last bit is 1 or 0. Number Mod 2 gives this result directly, as does Number And 1. Add this to the front of the string, then divide the number by 2, ignoring remainder.
13 Mod 2 = 1; Binary_String = "1"
13 \ 2 = 6; Loop
6 Mod 2 = 0; Binary_String = "01"
6 \ 2 = 3; Loop
3 Mod 2 = 1; Binary_String = "101"
3 \ 2 = 1; Loop
1 Mod 2 = 1; Binary_String = "1101"
1 \ 2 = 0; Exit Loop

Bin>>Dec coming up...

let me try:

Code:

---

Dim Number As Integer
Dim Binary_String As String
'copy the value of Text1 to Number
Number = Val(Text1.Text)
Binary_String = ""
'start a do while until Number is Zero
While Number > 0   
'add a new binary digit for each do loop starting from the right, the binary is 0 if 'the Number can be divided evenly by 2, or is 1 if it can not be divided evenly.
Binary_String = Str(Number Mod 2) & Binary_String   
'dived the number by 2 (Integer division)
Number = Number \ 2
'continue the loop
Wend
'put the Binary_string onto Text2.Text
Text2.Text = Binary_String

---

And the convert back I'd use soemthing like this

VB Code:

Dim Number As Integer
Dim Binary_String As String
Dim i As Integer
Number = 0
i = 1
Binary_String = Text2.Text
While Len(Binary_String) > 0
Number = Number + Val(Right(Binary_String, 1)) * i
Binary_String = Left(Binary_String, Len(Binary_String) - 1)
i = i * 2
Wend
Text1.Text = Number

Note: You need to change the code for the decimal to binary a bit, in ordr to get ridd of the spae characters
Use this line.

Code:

---

Binary_String = Trim(Str(Number Mod 2) & Binary_String)

---

VB Code:

Dim Binary_String As String
Dim DecimalValue As Long
Dim BitCount As Long
Dim i As Long
 
Binary_String = Text1.Text
BitCount = Len(Binary_String)
' Check length of string
If BitCount = 0 Then
  MsgBox "Please enter a binary string."
  Exit Sub
ElseIf BitCount > 31 Then
  MsgBox "String is too long"
  Exit Sub
End If
For i = 1 To BitCount
  ' Make sure all characters are "0" or "1"
  If Not Mid$(Binary_String, i, 1) Like "[01]" Then
    MsgBox "Text is not binary string"
    Exit Sub
  End If
  ' Add the value of each bit
  If Mid$(Binary_String, i, 1) = "1" Then
    DecimalValue = DecimalValue + 2 ^ (BitCount - i)
  End If
Next i
' Display result
MsgBox DecimalValue

ok i am gonna try it :) I'll let you know if it worked by me... tnx allready!

Code:

---

'Option Explicit
' Vbforums.com , By SESSI4ML
' Converts Decimal to Binary
' This is part of a larger program
'
Private Sub Form_Load()
    Text1.Text = gstrBinary
 
    Call Binary
End Sub

Private Sub Command1_Click()
    Call Binary
    Text1.SetFocus
End Sub
Sub Binary()
If InStr(1, Text2.Text, "|") Then
    Text1.Text = ""
    Text2.Text = ""
    Text1.SetFocus
    Exit Sub
    End If
Dim num As Double
If Text1.Text = "" Then Exit Sub
num = Text1.Text
Dim a As Double
Dim b As Double
Dim c As Double
Dim bin As String
Dim ss As String

    d% = 32
    Text3.SelStart = 0
    Text3.Text = ""
    Text4.Text = ""
aa:

    b = 2 ^ d%
    Text3.Text = Text3.Text + CStr(b) & vbCrLf
  If num > b Or b = num Then
        c = num / b
        bin = bin + "1"
        Text4.Text = Text4.Text + "1"
        num = num - b
        d% = d% - 1
        If d% = -1 Then
            Text2.Text = fBinary(bin)
            'Text4.Text = bin
            Exit Sub
        End If
    Else
        bin = bin + "0"
        Text4.Text = Text4.Text + "0"
        d% = d% - 1
        If d% = -1 Then
            Text2.Text = fBinary(bin)
          ' Text4.Text = bin
            Exit Sub
        End If
    End If
 GoTo aa
   

End Sub
Function fBinary(strBin As String) As String
    intlen = Len(strBin) + Int((Len(strBin) / 8))
    For a = 1 To intlen
        fBinary = fBinary + "0"
    Next a
    b = 0
    For a = Len(strBin) To 1 Step -1
          Mid(fBinary, intlen - b, 1) = Mid(strBin, a, 1)
              ct = ct + 1
            If ct = 8 Then
                b = b + 1
                ct = 0
                Mid(fBinary, intlen - b, 1) = "|"
            End If
            intlen = intlen - 1
    Next a
           
End Function

Private Sub Command2_Click()
    Unload Me
'  frmCalc.Show
End
End Sub


Private Sub Text1_KeyPress(KeyAscii As Integer)
  Select Case KeyAscii
        Case 8                      ' back space... delete last char
            ' Let char pass
    Case 48 To 57                  ' Digits 0-9
            ' Let char pass
     
    Case Else
        KeyAscii = 0
    End Select
End Sub 'Process_KeyBoard

---

Dec > Bin...VB File

If Not Mid$(Binary_String, i, 1) Like "[01]" Then....

what does this code do exactally?

If Mid$(Binary_String, i, 1) = "1" Then
DecimalValue = DecimalValue + 2 ^ (BitCount - i)

I meant this one :)

If the character doesn't match one of the charcters within the brackets Then...
For details about the Like operator

Hm? make up my mind! :D

If the character is "1" Then the bit is ON, so add the value of that bit to our decimal value. The value of each bit is 2 to the power of BitNumber, where BitNumber = 0 for the rightmost bit, and increases to the left. Since we are counting i from the left, we must subtract i from the length of the string to know how far it is from the right.

ah i get it, you really helped me! I'll set the thread to Resolved :D