CODEWITHSUNDEEP
c++
Diego Carrilho
Diego Carrilho

Reputation: 13

What's the best way to parse a packet data in c++?

I'm transcoding a Delphi App to C++ and got a problem to get the correct way to do the cases of packet headers in C++.

The Headers Comes as "Dword" in a UCHAR buf and i dont want make like: if (packet[0] == 0xFF && packet[1] == 0xFF...);

Ex: Delphi Working with switch case:

              case PDword(@Packet[6])^ of
                $68FB0200:
                  begin
                     //Dword match correctly
                  End;
              end;

is there a way to do the same in C++ like the Delphi Example?

Already checked for some methds but ever fails the comparsion.

UCHAR               *Packet ;
Packet = ParsePacket(buf->Buf, buf->Size);
// The ParsePacket Returns FB 00 00 00 78 00 01 F3 02 FD

DWORD Op1 = 0x01F302FD;
DWORD *p_dw = (DWORD*) Packet[6];
if (p_dw == Op1)
{
 //Dword never match...
}

Upvotes: 1

Views: 518

Answers (2)

Stephan Lechner
Stephan Lechner

Reputation: 35164

The obvious issue with DWORD *p_dw = (DWORD*) Packet[6] is that you take a value as an address. You probably meant DWORD *p_dw = (DWORD*)&(Packet[6]).

Note, however, that you get additional troubles for two reasons:

First, alignment could be wrong such that the pointer received is not a valid address to be taken for dereferencing a DWORD. This is undefined behaviour then.

Second, you depend on a specific endianess.

To overcome the first issue, you could use memcpy:

DWORD Op2 = 0;
memcpy(&Op2,Packet+6,sizeof(Op2));

To overcome both the first and the second issue, you'll need to sum the bytes on your own, e.g.:

DWORD Op2 = Packet[6]<<24 | Packet[7]<<16 | Packet[8]<<8 | Packet[9]

Upvotes: 0

Ben Voigt
Ben Voigt

Reputation: 283883

Yes, the portable C (and C++) way is:

DWORD const Op1 = 0x01F302FD;
if (0 == memcmp(Packet + 6, &Op1, sizeof Op1)) { ... }

Note that we haven't tried to access the bytes at Packet+6 as a 32-bit quantity... they're likely to be misaligned for that. We're just asking whether the same sequence of 4 bytes exists there as exist in Op1.

Upvotes: 2

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