On Mon, Jun 26, 2023 at 03:51:59PM +0200, Jan Kara wrote:
> On Mon 26-06-23 14:57:55, Ahelenia Ziemiańska wrote:
> > On Mon, Jun 26, 2023 at 02:19:42PM +0200, Ahelenia Ziemiańska wrote:
> > > > splice(2) differentiates three different cases:
> > > > if (ipipe && opipe) {
> > > > ...
> > > > if (ipipe) {
> > > > ...
> > > > if (opipe) {
> > > > ...
> > > >
> > > > IN_ACCESS will only be generated for non-pipe input
> > > > IN_MODIFY will only be generated for non-pipe output
> > > >
> > > > Similarly FAN_ACCESS_PERM fanotify permission events
> > > > will only be generated for non-pipe input.
> > Sorry, I must've misunderstood this as "splicing to a pipe generates
> > *ACCESS". Testing reveals this is not the case. So is it really true
> > that the only way to poll a pipe is a sleep()/read(O_NONBLOCK) loop?
> So why doesn't poll(3) work? AFAIK it should...
poll returns instantly with revents=POLLHUP for pipes that were closed
by the last writer.
Thus, you're either in a hot loop or you have to explicitly detect this
and fall back to sleeping, which defeats the point of polling:
-- >8 --
#define _GNU_SOURCE
#include <errno.h>
#include <fcntl.h>
#include <poll.h>
#include <stdio.h>
#include <unistd.h>
int main() {
char buf[64 * 1024];
struct pollfd pf = {.fd = 0, .events = POLLIN};
size_t consec = 0;
for (ssize_t rd;;) {
while (poll(&pf, 1, -1) <= 0)
;
if (pf.revents & POLLIN) {
while ((rd = read(0, buf, sizeof(buf))) == -1 && errno == EINTR)
;
fprintf(stderr, "\nrd=%zd: %m\n", rd);
}
if (pf.revents & POLLHUP) {
if (!consec++)
fprintf(stderr, "\n\tPOLLHUPs");
fprintf(stderr, "\r%zu", consec);
} else
consec = 0;
}
}
-- >8 --
And
-- >8 --
$ ./rdr < fifo
rd=12: Success
1779532 POLLHUPs
rd=5: Success
945087 POLLHUPs
rd=12: Success
^C
-- >8 --
corresponding to
-- >8 --
$ cat > fifo
abc
def
ghi
^D
$ echo zupa > fifo
$ cat > fifo
as
dsaa
asd
^C
-- >8 --
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