Hello,
Felix Wu, le mar. 18 juil. 2023 18:12:16 -0700, a ecrit:
> 02 == SYN so it looks good. But both tcpdump and wireshark (looking into
> packet
> dump provided by QEMU invocation)
Which packet dump?
> I added multiple prints inside slirp and confirmed the ipv6 version of [1] was
> reached.
> in tcp_output function [2], I got following print:
> qemu-system-aarch64: info: Slirp: AF_INET6 out dst ip =
> fdb5:481:10ce:0:8c41:aaff:fea9:f674, port = 52190
> qemu-system-aarch64: info: Slirp: AF_INET6 out src ip = fec0::105, port =
> 54322
> It looks like there should be something being sent back to the guest,
That's what it is.
> unless my understanding of tcp_output is wrong.
It looks so.
> To understand the datapath of guestfwd better, I have the following questions:
> 1. What's the meaning of tcp_input and tcp_output? My guess is the following
> graph, but I would like to confirm.
No, tcp_input is for packets that come from the guest, and tcp_output is
for packets that are send to the guest. So it's like that:
> tcp_input write_cb host send()
> QEMU --------> slirp -----------> QEMU --------------------> host
> <-------- <--------- <-----------------
> tcp_output slirp_socket_recv host recv()
> 2. I don't see port 6655 in the above process. How does slirp know 6655 is the
> port that needs to be visited on the host side?
Slirp itself *doesn't* know that port. The guestfwd piece just calls the
SlirpWriteCb when it has data coming from the guest. See the
documentation:
/* Set up port forwarding between a port in the guest network and a
* callback that will receive the data coming from the port */
SLIRP_EXPORT
int slirp_add_guestfwd(Slirp *slirp, SlirpWriteCb write_cb, void *opaque,
struct in_addr *guest_addr, int guest_port);
and
/* This is called by the application for a guestfwd, to provide the data to be
* sent on the forwarded port */
SLIRP_EXPORT
void slirp_socket_recv(Slirp *slirp, struct in_addr guest_addr, int guest_port,
const uint8_t *buf, int size);
Samuel
