On Wed, Aug 02, 2017 at 05:41:15AM -0700, jlu....@gmail.com wrote:
> I wrapped some code to do listen UDP port and set priority, like this
>
> func ListenUDP(t string, laddr *net.UDPAddr, priority int) (*net.UDPConn,
> error) {
> conn, err := net.ListenUDP(t, laddr)
[...]
> err = syscall.SetsockoptInt(int(file.Fd()), syscall.SOL_SOCKET,
> syscall.SO_PRIORITY, priority)
[...]
To cite the socket(7) manual page:
| SO_PRIORITY
| Set the protocol-defined priority for all packets to be sent on this
| socket. Linux uses this value to order the networking queues: packets
| with a higher priority may be processed first depending on the selected
| device queueing discipline. For ip(7), this also sets the IP
| type-of-service (TOS) field for outgoing packets.
So basically that means that on a supposedly congested link, or with a
busy receiving process which hardly keeps up, there may exist queues
of packets, and setting this option could make the kernel prefer packets
sent over this socket to be processed first -- if there exist other
sockets sending packets which end up in the same queue(s).
> And, read data from that port with timeout
>
> conn.SetReadDeadline(time.Now().Add(time.Second * 5))
> n, addr, err = conn.ReadFromUDP(buf)
>
> But, i found the deadline not work...
Hence, two questions:
- How exactly does it not work.
When you make such statement, it should go like this:
1) I did such and such preparations.
2) I observe such and such behaviour.
3) I would instead expect another behaviour; you then describe it.
- I doubt sessing SO_PRIORITY on a socket has something to do with read
deadline on it (which is merely a timeout which measures how much time
the read operation on a socket is allowed to take (and if it fails to
complete in the defined timespan, it's aborted).
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