Thank you for the information.
It solved your problem.
Regards.
Lokesh.
On Wed, 14 Aug 2024 17:57:40 +0530
Lokesh Chakka <lvenkatakumarchakka@gmail.com> wrote:
> hi stephen,
>
> Thanks for your response. However it seems the solution is not working. The
> following is the modified snippet....
>
> while( condition )
> {
> for( pkt_count=0; pkt_count<num_of_pkts_per_queue; pkt_count++ )
> {
> rte_mbuf_refcnt_set( mbuf[pkt_count], 2 );// setting to two -
> first and second iteration
> }
> for( pkt_count=0; pkt_count<5; pkt_count++ )
> {
> fprintf( stderr, "%s %d %u\n", __func__, __LINE__,
> rte_mbuf_refcnt_read( mbuf[pkt_count] ) );//able to print two- first and
> second iteration
> }
> if( rte_eth_tx_burst( port_id, 0, mbuf, num_of_pkts_per_queue ) !=
> num_of_pkts_per_queue )
> {
> fprintf( stderr, "%s %d %d %s\n", __func__, __LINE__, rte_errno,
> rte_strerror(rte_errno) );//failing second time
> rte_exit( EXIT_FAILURE, "%s %d rte_eth_tx_burst port id: %u\n",
> __func__, __LINE__, port_id );
> }
> fprintf( stderr, "%s %d port: %u sent %u packets\n", __func__,
> __LINE__, port_id, num_of_pkts_per_queue );//able to send once - first
> iteration only
> for( pkt_count=0; pkt_count<5; pkt_count++ )
> {
> fprintf( stderr, "%s %d %u\n", __func__, __LINE__,
> rte_mbuf_refcnt_read( mbuf[pkt_count] ) );//refcnt is printing one
> }
> }
>
> There seems to be some more gap in my understanding. Could you please help
> understand the issue?
>
> Thanks & Regards
> --
> Lokesh Chakka.
>
>
> On Mon, Aug 12, 2024 at 8:22 PM Stephen Hemminger <
> stephen@networkplumber.org> wrote:
>
> > On Mon, 12 Aug 2024 15:55:50 +0530
> > Lokesh Chakka <lvenkatakumarchakka@gmail.com> wrote:
> >
> > > hello,
> > >
> > > Here is a small piece of code :
> > >
> > > while( condition )
> > > {
> > >
> > > if( rte_eth_tx_burst( port_id, 0, mbuf, num_of_pkts_per_queue ) !=
> > > num_of_pkts_per_queue )
> > > {
> > > fprintf( stderr, "%d %s\n", rte_errno,
> > rte_strerror(rte_errno) );
> > > rte_exit( EXIT_FAILURE, "%s %d rte_eth_tx_burst port id:
> > %u\n",
> > > __func__, __LINE__, port_id );//second iteration failing.
> > > }
> > > fprintf( stderr, "%s %d port: %u packet: %c sent %u packets\n",
> > > __func__, __LINE__, port_id, argv[3][0], num_of_pkts_per_queue
> > > );//printing once
> > > for( pkt_count=0; pkt_count<num_of_pkts_per_queue; pkt_count++ )
> > > {//want to send same data again...!!!
> > > mbuf[pkt_count]->pkt_len = mbuf[pkt_count]->data_len =
> > dev_info.max_mtu;
> > >
> > > }
> > >
> > > }
> > >
> > > Can someone help me understand how to reuse the packets again to send the
> > > same data ?
> > >
> >
> > When packet is passed to tx_burst, the ownership of that mbuf is passed
> > to the driver. The driver will free it after it is sent.
> >
> > One option would be to increase the reference count on the packet before
> > sending.
> > Using rte_mbuf_refcnt_update() function to add one to refcount.
> > Then the driver will decrement refcount and the refcount will still be one
> > (not freed).
> >
> > Assume this is some kind of packet generator.
> >
Don't use refcnt_set(), you want to add an additional refcount not force it to two.
The device driver may hold onto the mbuf for a while after it was passed to tx_burst
so can't assume a value of two.
If the number actually queued is less that the number requested, that means the device
transmit queue is now full. Need to handle that case.
Something like this might get you started:
volatile running = true;
void flood(uint16_t port_id, struct rte_mbufs *mbufs[], uint16_t num_pkts)
{
while (running) {
/* Add additional reference to retain the packets */
for (uint16_t i = 0; i < num_pkts; i++)
rte_mbuf_refcnt_update(mbufs[i], 1);
uint16_t sent = rte_eth_tx_burst(port_id, queue_id, mbufs, num_pkts);
if (sent < num_pkts) {
/* device transmit queue was full, drop our reference */
for (uint16_t i = sent; i < num_pkts; i++)
rte_pktmbuf_free(mbufs[i]);
}
}
/* Don't leak original version */
rte_pktmbuf_free_bulk(mbufs, num_pkts);
}