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Thread: must be an easier way...[Resolved] figured it out myself!

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  1. #1

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    must be an easier way...[Resolved] figured it out myself!

    VB Code:
    1. Function DecimalToBinary(ByVal lngBin As Long, lngDesc As Long) As Long
    2.     Dim intNum As Integer
    3.    
    4.     Dim lngBinary2 As Long
    5.     Dim lngDecimal As Long
    6.    
    7.     intNum = txtNum.Text
    8.    
    9.     For intNum = intNum To 255
    10.         If intNum >= 2 ^ 7 Then
    11.             strNewNum = strNewNum & "1"
    12.             intNum = intNum - 2 ^ 7
    13.         Else
    14.             strNewNum = strNewNum & "0"
    15.         End If
    16.         If intNum >= 2 ^ 6 Then
    17.             strNewNum = strNewNum & "1"
    18.             intNum = intNum - 2 ^ 6
    19.         Else
    20.             strNewNum = strNewNum & "0"
    21.         End If
    22.         If intNum >= 2 ^ 5 Then
    23.             strNewNum = strNewNum & "1"
    24.             intNum = intNum - 2 ^ 5
    25.         Else
    26.             strNewNum = strNewNum & "0"
    27.         End If
    28.         If intNum >= 2 ^ 4 Then
    29.             strNewNum = strNewNum & "1"
    30.             intNum = intNum - 2 ^ 4
    31.         Else
    32.             strNewNum = strNewNum & "0"
    33.         End If
    34.         If intNum >= 2 ^ 3 Then
    35.             strNewNum = strNewNum & "1"
    36.             intNum = intNum - 2 ^ 3
    37.         Else
    38.             strNewNum = strNewNum & "0"
    39.         End If
    40.         If intNum >= 2 ^ 2 Then
    41.             strNewNum = strNewNum & "1"
    42.             intNum = intNum - 2 ^ 2
    43.         Else
    44.             strNewNum = strNewNum & "0"
    45.         End If
    46.         If intNum >= 2 ^ 1 Then
    47.             strNewNum = strNewNum & "1"
    48.             intNum = intNum - 2 ^ 1
    49.         Else
    50.             strNewNum = strNewNum & "0"
    51.         End If
    52.         If intNum >= 2 ^ 0 Then
    53.             strNewNum = strNewNum & "1"
    54.             intNum = intNum - 2 ^ 0
    55.         Else
    56.             strNewNum = strNewNum & "0"
    57.         End If
    58.         intNum = 256
    59.     Next intNum
    60. End Function

    there must be an easier way to do that without all those if statements... can some1 show me how plz?
    Last edited by MiddleAged1; Mar 10th, 2004 at 10:38 AM.
    I'm just a basic visual basic programmer!!! (no seriously... i'm just learning this stuff... the so called 'easy' stuff)... plz don't get mad at me if i don't know what i'm doing!!!

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