Question

I’m currently trying to get this C code converted into C#.
Since I’m not really familiar with C I’d really apprecheate your help!

static unsigned char byte_table[2080] = {0};

First of, some bytearray gets declared but never filled which I’m okay with

BYTE* packet = //bytes come in here from a file
int unknownVal = 0;
int unknown_field0 = *(DWORD *)(packet + 0x08);
do
{
    *((BYTE *)packet + i) ^= byte_table[(i + unknownVal) & 0x7FF];
    ++i;
}
while (i <= packet[0]);

But down here.. I really have no idea how to translate this into C#
BYTE = byte[] right?

DWORD = double?
but how can (packet + 0x08) be translated? How can I add a hex to a bytearray? Oo

I’d be happy about anything that helps! 🙂

Answer 1

In C, setting any set of memory to {0} will set the entire memory area to zeroes, if I’m not mistaken.

That bottom loop can be rewritten in a simpler, C# friendly fashion.

byte[] packet = arrayofcharsfromfile;
int field = packet[8]+(packet[9]<<8)+(packet[10]<<16)+(packet[11]<<24); //Assuming 32 bit little endian integer
int unknownval = 0;
int i = 0;

do //Why waste the newline? I don't know. Conventions are silly!
{
    packet[i] ^= byte_table[(i+unknownval) & 0x7FF];
} while( ++i <= packet[0] );

field is set by taking the four bytes including and following index 8 and generating a 32 bit int from them.

In C, you can cast pointers to other types, as is done in your provided snippet. What they’re doing is taking an array of bytes (each one 1/4 the size of a DWORD) and adding 8 to the index which advances the pointer by 8 bytes (since each element is a byte wide) and then treating that pointer as a DWORD pointer. In simpler terms, they’re turning the byte array in to a DWORD array, and then taking index 2, as 8/4=2.

You can simulate this behavior in a safe fashion by stringing the bytes together with bitshifting and addition, as I demonstrated above. It’s not as efficient and isn’t as pretty, but it accomplishes the same thing, and in a platform agnostic way too. Not all platforms are little endian.