[dpdk-users] Beginners question: rte_eth_tx_burst, rte_mbuf access synchronization

[Philipp Beyer]

Hi!

I am just writing my first code using dpdk, a traffic generator, for 
which I started with the l2fwd example.

Basically, I need to send the same packet over a single interface, over 
an over again, with single bytes changed each time.
I use rte_eth_tx_burst to send 16 packets at once. As I want to re-use 
the same buffers in a very simple way, I just increment the refcnt
accordingly.

My current code prepares all 16 buffers, calls rte_eth_tx_burst until 
all 16 packets are stored in the transmit ring, and starts over again, 
adjusting the buffers to send the next 16 packets.

Currently I observe duplicate packets, although every packet should be 
individual due to single byte adjustments.

My current problem is, as I guess, that rte_eth_tx_burst does not 
synchnolously transmit the count of packets, which is returned to the 
caller, but just stores them in transmit queue. So, I am not allowed to 
instantly re-use these buffers again.

My question is: How do I know when to re-use buffers passed to 
rte_eth_tx_burst. Of course, I can check their refcnt member, and this 
would be perfectly fine. Apparently, I should have at least BURST_SIZE*2 
buffers, passing BURST_SIZE buffers at once, so I can manipulate one set 
of buffers while the other is transmitted. But I am missing the idea of 
the best synchronization scheme here: How should I wait on this refcnt 
to drop?

Some blind guessing:
If I take the documentation of rte_eth_tx_burst literally, I could get 
the idea that refcounts of buffers are only decreased (buffers are 
'freed'), while rte_eth_tx_burst is executed, but one function call 
might free buffers used by previous function calls. If this is correct, 
I still do not see a complete synchronization scheme. There is still a 
chance that I end up without any buffers left, which means I do not have 
a chance to call rte_eth_tx_burst again to free buffers.

Thanks for any help,
Philipp