David Miller <davem@xxxxxxxxxxxxx> writes:
> From: Rainer Weikusat <rweikusat@xxxxxxxxxxxxxxxxxxxxxxx>
> Date: Mon, 16 Nov 2015 22:28:40 +0000
>
>> An AF_UNIX datagram socket being the client in an n:1
[...]
> So because of a corner case of epoll handling and sender socket release,
> every single datagram sendmsg has to do a double lock now?
>
> I do not dispute the correctness of your fix at this point, but that
> added cost in the fast path is really too high.
Some more information on this: Running the test program included below
on my 'work' system (otherwise idle, after logging in via VT with no GUI
running)/ quadcore AMD A10-5700, 3393.984 for 20 times/ patched 4.3 resulted in the
following throughput statistics[*]:
avg 13.617 M/s
median 13.393 M/s
max 17.14 M/s
min 13.047 M/s
deviation 0.85
I'll try to post the results for 'unpatched' later as I'm also working
on a couple of other things.
[*] I do not use my fingers for counting, hence, these are binary and
not decimal units.
------------
#include <inttypes.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/socket.h>
#include <sys/time.h>
#include <sys/wait.h>
#include <unistd.h>
enum {
MSG_SZ = 16,
MSGS = 1000000
};
static char msg[MSG_SZ];
static uint64_t tv2u(struct timeval *tv)
{
uint64_t u;
u = tv->tv_sec;
u *= 1000000;
return u + tv->tv_usec;
}
int main(void)
{
struct timeval start, stop;
uint64_t t_diff;
double rate;
int sks[2];
unsigned remain;
char buf[MSG_SZ];
socketpair(AF_UNIX, SOCK_SEQPACKET, 0, sks);
if (fork() == 0) {
close(*sks);
gettimeofday(&start, 0);
while (read(sks[1], buf, sizeof(buf)) > 0);
gettimeofday(&stop, 0);
t_diff = tv2u(&stop);
t_diff -= tv2u(&start);
rate = MSG_SZ * MSGS;
rate /= t_diff;
rate *= 1000000;
printf("rate %fM/s\n", rate / (1 << 20));
fflush(stdout);
_exit(0);
}
close(sks[1]);
remain = MSGS;
do write(*sks, msg, sizeof(msg)); while (--remain);
close(*sks);
wait(NULL);
return 0;
}
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