David Miller [off-list ref] writes:

From: Rainer Weikusat <redacted>
Date: Mon, 16 Nov 2015 22:28:40 +0000

quoted

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;
}