Re: read -t 0 anomaly

[Pierre Gaston]

On Fri, Oct 4, 2013 at 3:08 PM, Kunszt Árpád <
arpad.kunszt@syrius-software.hu> wrote:

> On 2013. October 4. 14:51:00 Pierre Gaston wrote:
> > On Fri, Oct 4, 2013 at 2:20 PM, Kunszt Árpád
> ...
> >
> >
> > There is a race condition, you cannot know if echo will run before read.
>
> I see, and it's logical. But this stills confuses me.
>
> arpad@terminus ~ $ for(( i=0; i<10; i++ )); do echo -n "" | { sleep 1s;
> read -t 0; echo $?; } ; done | sort | uniq -c
>      10 0
>
> I expected that the read will return non-zero in this case. I think it
> returned with zero because the STDIN was still open. The docs said "read
> returns success if input is available on the specified file descriptor,
> failure otherwise". There wasn't any data on the file descriptor, it was
> just open.
>
> Am I still doing something wrong? Or I just misunderstanding the meaning
> of "input is available" term? I'm not a native English speaker (as you can
> se from my mails clearly).
>
> Thanks,
>
> Arpad Kunszt
>
>
Most probably read uses and does what the select() call does.
In my man select(2) I have:

"more  precisely, to see  if a  read will  not block;  in particular,  a
file  descriptor is  also  ready  on  end-of-file"

and that's what your exemple does, echo opens stdin and then closes it and
read sees and end of file.


no worry about your english..