vb vs c xoring

[shardic]

I need to convert the following C code to VB such that the resulting bit configurations are identical. Note that I cannot alter the variable types.

Being relatively new to VB I am a little confused concerning XOR'in VB ints with chars. Any advice is greatly appreciated.

char *somefunc( char *instr, char *pwstr, long len )
{
int j, pwlen, passnum;
long i;
char buff[1];
pwlen = strlen( pwstr ); /* length of password */
/* get seed value */
passnum = (int) ((((_tr_pnum(pwstr)/997) - 1) % 254 ) + 1);
for ( i = j = 0; i < len; i++ ) /* process whole string */
{
passnum = (int) (((passnum + ( i - len )) - 1 ) % 254) + 1;
buff[0] = (instr ^ (passnum ^ pwstr[j]) ); /* XOR 3 var's */
instr = (buff[0] ? buff[0] : instr); /* if NULL retrn char */
j = ( j >= pwlen ? 0 : j++ ); /* password que control variable */
}

return(instr); /* send back encrypted string */
}

 

[vis781]

Do While i < len ' process whole string 
passnum = CInt((passnum + (i - len)) - 1) Mod 254) + 1
buff(0) = (instr(i) Xor (passnum Xor pwstr(j))) ' XOR 3 var's 
If buff(0) Then
 instr(i) = (buff(0))
Else
 instr(i) = (instr(i))
End If
If j >= pwlen Then
 j = (0)
Else
 j += 1
End If
 i += 1
Loop

Well, above is my effort at converting that loop. This is where c# differs greatly from VB.Net. I haven't tested it, just done a syntactical translation. Let me know how it goes.

 

[shardic]

Thanks for the reply. This still has the issue of 'xor not defined for types Integer and char' for (passnum Xor pwstr(j))
This is where I get confused concerning the bit configuration if I convert the int to a char or the char to an int.

 

[vis781]

As i say i haven't been able to test it out myself. I'm assuming that you have tried to convert the values? I'd use the BitConverter Class to do it.

 

[shardic]

Thank you for that "little" clue. I have not been aware of the bitconverter class and that seems to be exactly what I needed. The process of converting all entities to BYTE and then xor'ing seems to yield the same results as executing the C program.

Again, thanks for your help.