System.Byte() array convert to string... HOW???

[Jarrod Sharp]

Using the VB6 Winsock control in VB.NET the GetData(byRef Data as Object) method of the control returns an Object.

I have used this code that works but am not familiar with .NET arrays/object/conversions...

Private Sub wskSocketPlus_D ataArrival(ByVa l sender As Object, ByVal e As AxMSWinsockLib. DMSWinsockContr olEvents_DataAr rivalEvent) Handles wskSocketPlus.D ataArrival

Dim o As Object
wskSocketPlus.G etData(o)
Dim incomingDataArr ay() As Byte
incomingDataArr ay = CType(o, System.Byte())

Dim Buffer As New StringBuilder
Dim b As Byte

For element As Integer = 0 To incomingDataArr ay.Length - 1
b = Convert.ToByte( incomingDataArr ay.GetValue(ele ment))
Buffer.Append(C hr(b))
Next element

MessageBox.Show (Buffer.ToStrin g)

End Sub

I was told I should use string builder as it is fatser than concatination. The datapackets are around a few thosand bytes. Any suggestions as to a more correct set of conversions to give me a resulting string of the incoming Object data?

thanks, Jarrod

[Jarrod Sharp]

RE: System.Byte() array convert to string... HOW???

this worked as well.... but geez.... why no GetData(string) ??

Dim o As Object
AxWinsock2.GetD ata(o)

Dim b() As Byte
b = CType(o, Byte())

Dim a As ASCIIEncoding
a = New ASCIIEncoding

Dim s As String
s = a.GetString(b)

MessageBox.Show (s)

[Jon Skeet [C# MVP]]

RE: System.Byte() array convert to string... HOW???

Jarrod Sharp <JarrodSharp@di scussions.micro soft.com> wrote:[color=blue]
> this worked as well.... but geez.... why no GetData(string) ??[/color]

Because a socket is a fundamentally binary object. I think it's great
that the framework makes such a clear distinction between binary data
and character data.

Note that you don't need to create a new instance of ASCIIEncoding
yourself - just use the Encoding.ASCII property.

(I'd also suggest using the Socket class that's built into the
framework, btw - or even TcpClient.)

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too