Help With Porting Code From C++

[teK]

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unsigned char *GamePacketSize(unsigned char *data, unsigned int *size,
                            unsigned int *offset)
{
  unsigned int a;

  if (data[0] < 0xF0) {
      *size = data[0] - 1;
      *offset = 1;
      return &data[1];
  }

  a = (data[0] & 0xF) << 8;
  *size = a + data[1] - 2;
  *offset = 2;
  return &data[2];
}


Anyone mind porting/help porting this to VB?

Thanks.

[CrAz3D]

I'm sure there is some awesome tutorial that you can find through google, I mean google pwns my mom even!

[UserLoser.]

Anyone mind porting/help porting this to VB?

Thanks.

I believe that code is old and outdated, but anyways shouldn't be hard at all

[teK]

Anyone mind porting/help porting this to VB?

Thanks.

I believe that code is old and outdated, but anyways shouldn't be hard at all

Never knew that , oh well. Anyone know where I can find an updated version?

[UserLoser.]

Anyone mind porting/help porting this to VB?

Thanks.

I believe that code is old and outdated, but anyways shouldn't be hard at all

Never knew that , oh well. Anyone know where I can find an updated version?

Since you havn't told anyone what it's for, I will.  This function is used to calculate the size of a D2GS compressed packet, from the server before it's actually decompressed.  I'm not exactly sure if that's working or not, but IIRC, the code someone posted here a while back (which is that) didn't work properly.  Also, you can find it in D2Net.dll

[LivedKrad]

Let me see...

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Private/Public Function GamePacketSize (ByRef data() As Integer?, ByVal size as Integer,ByVal offset As Integer) As Integer?

Dim a As Integer

If data(0) < &HF Then
size = data(0) - 1
offset = 1
GamePacketSize = data(1)
End If

((((a = (data[0] & 0xF) << 8;))) Some sort of shift operation?
size = a + data(1) - 2
offset = 2
GamePacketSize = data(2)



Hope that helps at all =(

Edit: Changed "GamePacketSize = data[1]" to (1)

[drivehappy]

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Private Function GamePacketSize (ByRef data() As Byte, ByRef size As Integer, ByRef offset As Integer) As Byte
   Dim a As Integer

    If data(0) < &HF0 Then
        size = data(0) - 1
        offset = 1
        GamePacketSize = data(1)
    Else
         a = ((data(0) And &HF) * 256)
         size = a + data(1) - 2
         offset = 2
         GamePacketSize = data(2)
    End If

End Function


You will probably want a shift left function instead of multiplying it by 256 because of overflow errors if the data is too large.

[UserLoser.]

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    If data(0) < &HF Then


You will probably want a shift left function instead of multiplying it by 256 because of overflow errors if the data is too large.

Note: both you and LivedKrad failed to realize it's 0xF0, not 0xF.

[LivedKrad]

Hmm, then it would seem Grok's explanation in Op [vL] was inaccurate. I believe he explained about creating a new string with the data located at

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data[0] plus 1111. According to this however, 1111 is wrong. I believe 0xF0 is 240, not 15. Sorry for the confusion [original poster]. I don't know C++ so if there were some inconsistencies within the code I'm sorry. And, before UserLoser says it, I'm not saying mistaking 0xF and 0xF0 was concerning knowledge of C++. That was my mistake,

[Grok]

Hmm, then it would seem Grok's explanation in Op [vL] was inaccurate. I believe he explained about creating a new string with the data located at

Code Select Expand

data[0] plus 1111. According to this however, 1111 is wrong. I believe 0xF0 is 240, not 15. Sorry for the confusion [original poster]. I don't know C++ so if there were some inconsistencies within the code I'm sorry. And, before UserLoser says it, I'm not saying mistaking 0xF and 0xF0 was concerning knowledge of C++. That was my mistake,

Wow, blaming me?

[13:07:40] <LivedKrad> still makes no sense what that operation is: a = (data[0] & 0xF) << 8;

I responded in the channel the code you posted in the channel.

[LivedKrad]

Well, guess it's my fault then. Sorry.

[drivehappy]

Code Select Expand


    If data(0) < &HF Then


You will probably want a shift left function instead of multiplying it by 256 because of overflow errors if the data is too large.

Note: both you and LivedKrad failed to realize it's 0xF0, not 0xF.

Fixed. I was trying to save typing and copied both LivedKrad's and the original C++ code into the box. Goes to show shortcutting doesn't pay, plus the post message textbox is too small vertical wise so I can't see all of the code.

[l)ragon]

a = ((data(0) And &HF)  * (2 ^ 8))

Edit: dont use 256 the way that you are 9p

[drivehappy]

a = ((data(0) And &HF)  * (2 ^ )

Edit: dont use 256 the way that you are 9p

Why not? Last time I checked 2^8 = 256

[l)ragon]

a = ((data(0) And &HF)  * (2 ^ 8))

Edit: dont use 256 the way that you are 9p

Why not? Last time I checked 2^8 = 256

test which one breaks then lol bottom one or the top one

dim x as long, x2 as long
x = 255 * 256
x2 = 255 * (2 ^ 8)

edit: typo
edit2: yes 2^8 = 256