[Xorp-hackers] [PIM] Setting a bit in the Encoded source address

[Pavlin Radoslavov]

chintamani wandhre <chintamanisw at gmail.com> wrote:

> hello,
> We have to set the bidirectional bit in the ENCODED GROUP ADDRESS
> 
> 
>  0                   1                   2                   3
>  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
> |  Addr Family  | Encoding Type |B| Reserved  |Z|  Mask Len     |
> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
> |                Group multicast Address
> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+...
> 
> By using following function we can get octet but how can we set the single
> bit ( B )?  :)
> 
> i could not get the flow in the following code :
> 
> #define GET_OCTET(val, cp, rcvlen)                      \
> do {                                                                    \
>     if ((size_t)(rcvlen) < (size_t)1)                            \
>         goto rcvlen_error;                                         \
>     (rcvlen)--;                                                         \
>     ((val) = *(cp)++);                                               \
> } while (0)
> 
> thanking you in anticipation

You could use the PUT_OCTET() macro.
I believe the remaining 7 bits in the Reserved field are suppose to
be 0. Hence you need to use PUT_OCTET() to set both Reserved and Z
at the same time (because they belong to the same octet).

Regards,
Pavlin