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disable stdout buffering ?

P: n/a
Hi.

I have the following problem:
Using popen() to execute a program and read its stdout works usually fine.
Now I try to do this with a program called xsupplicant (maybe one knows),
but I dont get the output of it while it is running.

This is probably a problem of stdout being buffered, because if I use
fflush() after a printf() in the xsupplicant then I can read the output.

My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.

I dont want to change the xsupplicant source code.

Hope anyone understood what I am trying to say.
thanks.
Nov 15 '05 #1
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7 Replies


P: n/a
>
My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.

Yes. Please see setvbuf and other related functions.

Regards,
Madhav.

Nov 15 '05 #2

P: n/a
Thanks for your reply.

On Tue, 04 Oct 2005 05:50:26 -0700, Madhav wrote:

My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.

Yes. Please see setvbuf and other related functions.

I tried setvbuf and setbuf:

....
FILE* fp = popen("...","r");
setvbuf(fp,NULL,_IONBF,0);
....

But unfortunately it had no effect ...

Nov 15 '05 #3

P: n/a
Mathias Herrmann <herrmann@'remove_me'sit.fraunhofer.de> wrote:
On Tue, 04 Oct 2005 05:50:26 -0700, Madhav wrote:
My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.

Yes. Please see setvbuf and other related functions.


I tried setvbuf and setbuf:

...
FILE* fp = popen("...","r");
setvbuf(fp,NULL,_IONBF,0);


Ah. You are dealing with pipes, not with normal streams. Pipes are
different; they may or may not respond to functions that work on files.
Pipes are also not part of ISO C, and different implementations of pipes
may behave in different ways. For a reliable answer, ask in a newsgroup
for whatever library defines your pipes - probably POSIX, in which case,
ask in comp.unix.programmer.

Richard
Nov 15 '05 #4

P: n/a


Mathias Herrmann wrote On 10/04/05 09:22,:
Thanks for your reply.

On Tue, 04 Oct 2005 05:50:26 -0700, Madhav wrote:

My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.


Yes. Please see setvbuf and other related functions.


I tried setvbuf and setbuf:

...
FILE* fp = popen("...","r");
setvbuf(fp,NULL,_IONBF,0);
...

But unfortunately it had no effect ...


... because it was done on "the wrong end" of the
connection. You need to disable the buffering that
the sender is doing, so you need to get the sender to
call setvbuf() before doing anything else with its
stdout stream.

Unfortunately, that goes against your goal of not
wanting to make any changes to the other program's code.
However, the change is a very small one: just a one-liner
at or near the start of main() will do it.

Unfortunately (again), you're still at the mercy of
the operating system. Telling the C library not to buffer
an output stream causes fprintf() and the like to deliver
characters directly to the O/S without undue delay, but
doesn't govern what the O/S then chooses to do with them.
You'll need to read your O/S' documentation on pipes (which
aren't part of C, by the way).

<off-topic>

In my experience, many "interactive" programs fare
rather poorly with pipes; the I/O model really isn't quite
what's wanted. If your system provides pseudo-terminals
(/dev/pty), they might be more appropriate. For further
help along these lines -- or with pipes, for that matter --
try comp.unix.programmer.

</off-topic>

--
Er*********@sun.com

Nov 15 '05 #5

P: n/a
> ... because it was done on "the wrong end" of the
connection. You need to disable the buffering that
the sender is doing, so you need to get the sender to
call setvbuf() before doing anything else with its
stdout stream.

Unfortunately, that goes against your goal of not
wanting to make any changes to the other program's code.
However, the change is a very small one: just a one-liner
at or near the start of main() will do it.

Unfortunately (again), you're still at the mercy of
the operating system. Telling the C library not to buffer
an output stream causes fprintf() and the like to deliver
characters directly to the O/S without undue delay, but
doesn't govern what the O/S then chooses to do with them.
You'll need to read your O/S' documentation on pipes (which
aren't part of C, by the way).

<off-topic>

In my experience, many "interactive" programs fare
rather poorly with pipes; the I/O model really isn't quite
what's wanted. If your system provides pseudo-terminals
(/dev/pty), they might be more appropriate. For further
help along these lines -- or with pipes, for that matter --
try comp.unix.programmer.

</off-topic>

Thanks a lot.
Im going to try the use of a pty.
Nov 15 '05 #6

P: n/a
Mathias Herrmann wrote:

Hi.

I have the following problem:
Using popen() to execute a program and read its stdout works usually fine.
Now I try to do this with a program called xsupplicant (maybe one knows),
but I dont get the output of it while it is running.

This is probably a problem of stdout being buffered, because if I use
fflush() after a printf() in the xsupplicant then I can read the output.

My question is now: is it possible to tell stdout not to buffer anything,
but immediately print it out.

I dont want to change the xsupplicant source code.


You can't "unbuffer" xsupplicant's stdout without changing xsupplicant's
source code, unless this ability is already built in.

You could probably try something like this at the start of main():

if ( !isatty(stdout) )
setbuf(stdout,NULL);

The specifics are probably OT to clc, so you might need to ask in a
group where this would be on-topic. Perhaps comp.unix.programmer?

--
+-------------------------+--------------------+-----------------------------+
| Kenneth J. Brody | www.hvcomputer.com | |
| kenbrody/at\spamcop.net | www.fptech.com | #include <std_disclaimer.h> |
+-------------------------+--------------------+-----------------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>
Nov 15 '05 #7

P: n/a

In article <pan.2005.10.04.12.11.37.993986@'remove_me'sit.fra unhofer.de>, Mathias Herrmann <herrmann@'remove_me'sit.fraunhofer.de> writes:

My question is now: is it possible to tell stdout not to buffer anything,
Yes:

#include <stdio.h>

int main(void)
{
int ret;
ret = setvbuf(stdout, NULL, _IONBF, 0);
if (ret == 0)
puts("stdout is now unbuffered");
return 0;
}
but immediately print it out.


No. While you can disable the C library's buffering for stdout
(if setvbuf succeeds), you don't have any control over when the
underlying system processes the output generated by the C library.

However, I believe you've asked the wrong question, since you
appear to want to change the stdout buffering for another process,
without changing that program's source code. That you cannot do,
in standard C. (Note that popen is not part of the standard C
library.)

I suggest you take this question to a newsgroup that deals with
popen, such as comp.unix.programmer, or one specific to your
operating system. (OT: Unless I am gravely mistaken, there's no
POSIX/SUS mechanism to force another process to disable stdio
buffering, either. Using a pseudo-tty rather than a pipe should
get you line buffering rather than full buffering, though, which
might suffice.)
--
Michael Wojcik mi************@microfocus.com

You brung in them two expert birdwatchers ... sayin' it was to keep us from
makin' dern fools of ourselfs ... whereas it's the inherent right of all to
make dern fools of theirselfs ... it ain't a right held by you official types
alone. -- Walt Kelly
Nov 15 '05 #8

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