Newbie Question about multiple system calls within loops

Discussion in 'C Programming' started by student.matt@gmail.com, Jun 26, 2008.

1. Guest

ok i am trying to solve this little problem i have my program waits
for the 1st system() call to finish before starting the next i will
just throw an example of what i mean out

in this example it waits for the 1st kedit to be closed before it
opens the second

#include <stdio.h>
#include <stdlib.h>
int main()
{

int count = 0;

while(count <= 5)
{
system("kedit");
count++;
}

return 0;

}

what i would like to figure out is to make all 5 kedits open at once,
once agian i am fairly new i am just trying to figure out basic
concepts and in this one why exactly it waits for the 1st call to
finish before making the second

, Jun 26, 2008

2. rahulGuest

On Jun 26, 8:01 am, wrote:
> ok i am trying to solve this little problem i have my program waits
> for the 1st system() call to finish before starting the next i will
> just throw an example of what i mean out

You should not be using the term system call when what you actually
mean is invoking the system library function.
> in this example it waits for the 1st kedit to be closed before it
> opens the second

Did you consider looking at the documentation for system() ?

NAME
system - execute a shell command

SYNOPSIS
#include <stdlib.h>

int system(const char *command);

DESCRIPTION
system() executes a command specified in command by calling /
bin/sh -c command, and returns after the
command has been completed. During execution of the command,
SIGCHLD will be blocked, and SIGINT and
SIGQUIT will be ignored.

I am posting Linux man pages but the basic idea holds true for any
platform. system() will not return until and unless the command you
have invoked through system is completed.

rahul, Jun 26, 2008

3. Johannes BauerGuest

schrieb:

> what i would like to figure out is to make all 5 kedits open at once,
> once agian i am fairly new i am just trying to figure out basic
> concepts and in this one why exactly it waits for the 1st call to
> finish before making the second

Have a look at fork/exec.

Note that this question better belongs to
comp.os.linux.development.system (or the system you're working with).
You're probably gonna get yelled at here because this group deals with
the C language (not the system programming in C).

Regards,
Johannes

--
"Wer etwas kritisiert muss es noch lange nicht selber besser können. Es
reicht zu wissen, daß andere es besser können und andere es auch
besser machen um einen Vergleich zu bringen." - Wolfgang Gerber
in de.sci.electronics <47fa8447$0$11545\$>

Johannes Bauer, Jun 26, 2008
4. Stephen SprunkGuest

wrote:
> ok i am trying to solve this little problem i have my program waits
> for the 1st system() call to finish before starting the next i will
> just throw an example of what i mean out
>
> in this example it waits for the 1st kedit to be closed before it
> opens the second
>
> #include <stdio.h>
> #include <stdlib.h>
> int main()
> {
>
> int count = 0;
>
> while(count <= 5)
> {
> system("kedit");
> count++;
> }
>
> return 0;
>
> }
>
>
> what i would like to figure out is to make all 5 kedits open at once,
> once agian i am fairly new i am just trying to figure out basic
> concepts and in this one why exactly it waits for the 1st call to
> finish before making the second

By definition, system() does not return until the command has been
completed.

Your system may have some way to "detach" a shell command so that
system() will return immediately (e.g. "kedit &"), but that's inherently
non-portable. Likewise, it may have other ways to do the same thing
without system(), e.g. fork()/exec(), but those are also non-portable.

If you are willing to restrict your program to POSIX environments, then
you may wish to take advantage of some of the additional functionality
available from the POSIX (not C) standard, but that's off-topic here.

S

Stephen Sprunk, Jun 26, 2008
5. Guest

On Jun 26, 6:02 pm, Stephen Sprunk <> wrote:
> wrote:
> > ok i am trying to solve this little problem i have my program waits
> > for the 1st system() call to finish before starting the next i will
> > just throw an example of what i mean out

>
> > in this example it waits for the 1st kedit to be closed before it
> > opens the second

>
> > #include <stdio.h>
> > #include <stdlib.h>
> > int main()
> > {

>
> > int count = 0;

>
> > while(count <= 5)
> > {
> > system("kedit");
> > count++;
> > }

>
> > return 0;

>
> > }

>
> > what i would like to figure out is to make all 5 kedits open at once,
> > once agian i am fairly new i am just trying to figure out basic
> > concepts and in this one why exactly it waits for the 1st call to
> > finish before making the second

>
> By definition, system() does not return until the command has been
> completed.

By whose definition? Certainly not ISO 9899:1999's.

> Your system may have some way to "detach" a shell command so that
> system() will return immediately (e.g. "kedit &"), but that's inherently
> non-portable. Likewise, it may have other ways to do the same thing

system("kedit &") is as portable as system("kedit")

> without system(), e.g. fork()/exec(), but those are also non-portable.
>
> If you are willing to restrict your program to POSIX environments, then
> you may wish to take advantage of some of the additional functionality
> available from the POSIX (not C) standard, but that's off-topic here.

Agreed. It's topical in comp.unix.programmer, so OP can try that group
for POSIX questions.

, Jun 26, 2008
6. Guest

On Jun 26, 5:34 pm, wrote:
> On Jun 26, 6:02 pm, Stephen Sprunk <> wrote:
>
> > wrote:
> > > ok i am trying to solve this little problem i have my program waits
> > > for the 1st system() call to finish before starting the next i will
> > > just throw an example of what i mean out

>
> > > in this example it waits for the 1st kedit to be closed before it
> > > opens the second

>
> > > #include <stdio.h>
> > > #include <stdlib.h>
> > > int main()
> > > {

>
> > > int count = 0;

>
> > > while(count <= 5)
> > > {
> > > system("kedit");
> > > count++;
> > > }

>
> > > return 0;

>
> > > }

>
> > > what i would like to figure out is to make all 5 kedits open at once,
> > > once agian i am fairly new i am just trying to figure out basic
> > > concepts and in this one why exactly it waits for the 1st call to
> > > finish before making the second

>
> > By definition, system() does not return until the command has been
> > completed.

>
> By whose definition? Certainly not ISO 9899:1999's.
>
> > Your system may have some way to "detach" a shell command so that
> > system() will return immediately (e.g. "kedit &"), but that's inherently
> > non-portable. Likewise, it may have other ways to do the same thing

>
> system("kedit &") is as portable as system("kedit")
>
> > without system(), e.g. fork()/exec(), but those are also non-portable.

>
> > If you are willing to restrict your program to POSIX environments, then
> > you may wish to take advantage of some of the additional functionality
> > available from the POSIX (not C) standard, but that's off-topic here.

>
> Agreed. It's topical in comp.unix.programmer, so OP can try that group
> for POSIX questions.

Thanks for all the responses sorry for posting here, but the
system("kedit &"); did work read the man on system() little bit better
understanding of whats going on. i looked up 10 different
documentations on fork/exec it seems a little complicated just yet

but thanks for the help

, Jun 26, 2008
7. Antoninus TwinkGuest

On 26 Jun 2008 at 19:13, wrote:
>> > wrote:
>> > > what i would like to figure out is to make all 5 kedits open at once,

>
> Thanks for all the responses sorry for posting here, but the
> system("kedit &"); did work read the man on system() little bit better
> understanding of whats going on. i looked up 10 different
> documentations on fork/exec it seems a little complicated just yet

It's really not all that complicated. Here's a simple example that

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>

int main(void)
{
pid_t pid;
int count;
for(count=0; count<5; count++) {
pid=fork();
if (pid==0) {
/* child process */
execlp("kedit", "kedit", (char *) NULL);
perror("exec failure");
exit(1);
}
}
return 0;
}

Antoninus Twink, Jun 26, 2008

On Jun 26, 12:28 pm, Antoninus Twink <> wrote:
> On 26 Jun 2008 at 19:13, wrote:
>
> >> > wrote:
> >> > > what i would like to figure out is to make all 5 kedits open at once,

>
> > Thanks for all the responses sorry for posting here, but the
> > system("kedit &"); did work read the man on system() little bit better
> > understanding of whats going on. i looked up 10 different
> > documentations on fork/exec it seems a little complicated just yet

>
> It's really not all that complicated. Here's a simple example that
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <unistd.h>
> #include <sys/types.h>
>
> int main(void)
> {
>   pid_t pid;
>   int count;
>   for(count=0; count<5; count++) {
>     pid=fork();
>     if (pid==0) {
>       /* child process */
>       execlp("kedit", "kedit", (char *) NULL);

Maybe I misunderstood section 8.10 in the book "Advance Programming in
the Unix Environment" by Stevens and Rago, but I believe
(char *)NULL actually creates undefined behavior when it's passed to
the exec() family.

>       perror("exec failure");
>       exit(1);
>     }
>   }
>   return 0;
>

I thought it was supposed to be exit(0) and not

return 0

in this case.
>
>
> }- Hide quoted text -
>

On Jun 26, 12:28 pm, Antoninus Twink <> wrote:
> On 26 Jun 2008 at 19:13, wrote:
>
> >> > wrote:
> >> > > what i would like to figure out is to make all 5 kedits open at once,

>
> > Thanks for all the responses sorry for posting here, but the
> > system("kedit &"); did work read the man on system() little bit better
> > understanding of whats going on. i looked up 10 different
> > documentations on fork/exec it seems a little complicated just yet

>
> It's really not all that complicated. Here's a simple example that
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <unistd.h>
> #include <sys/types.h>
>
> int main(void)
> {
>   pid_t pid;
>   int count;
>   for(count=0; count<5; count++) {
>     pid=fork();
>     if (pid==0) {
>       /* child process */
>       execlp("kedit", "kedit", (char *) NULL);
>       perror("exec failure");
>       exit(1);
>     }
>   }
>   return 0;
>
>
>
> }- Hide quoted text -
>
> - Show quoted text -

And off course this is again off topic. So not only does this troll
fail to understand that this is a C forum, but this person writes
questionable simple *nix code.

10. Keith ThompsonGuest

> On Jun 26, 12:28 pm, Antoninus Twink <> wrote:

[...]
> > #include <stdio.h>
> > #include <stdlib.h>
> > #include <unistd.h>
> > #include <sys/types.h>
> >
> > int main(void)
> > {
> > pid_t pid;
> > int count;
> > for(count=0; count<5; count++) {
> > pid=fork();
> > if (pid==0) {
> > /* child process */
> > execlp("kedit", "kedit", (char *) NULL);

>
> Maybe I misunderstood section 8.10 in the book "Advance Programming in
> the Unix Environment" by Stevens and Rago, but I believe
> (char *)NULL actually creates undefined behavior when it's passed to
> the exec() family.

<OT>No, it's correct; it's used to mark the end of the arguments.</OT>

> > perror("exec failure");
> > exit(1);
> > }
> > }
> > return 0;
> >

> I thought it was supposed to be exit(0) and not
>
> return 0
>
> in this case.

Within main(), exit(0)'' and return 0'' are almost exactly
equivalent.

exit(1) is non-portable (the only portable exit arguments are 0,
EXIT_SUCCESS, and EXIT_FAILURE), but the code is non-portable anyway.

--
Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

Keith Thompson, Jun 27, 2008
11. Martien VerbruggenGuest

On Thu, 26 Jun 2008 17:20:50 -0700 (PDT),
> On Jun 26, 12:28 pm, Antoninus Twink <> wrote:
>> On 26 Jun 2008 at 19:13, wrote:

>>       execlp("kedit", "kedit", (char *) NULL);

>
> Maybe I misunderstood section 8.10 in the book "Advance Programming in
> the Unix Environment" by Stevens and Rago, but I believe
> (char *)NULL actually creates undefined behavior when it's passed to
> the exec() family.

\begin{offtopic}

I think you've got it the wrong way around. The last argument in the
variable list _has_ to be cast to char *.

This is, of course, not a standard C function, but a POSIX one, and
therefore it'd be better to discuss its interface on
comp.unix.programmer, rather than here. It is probably a good idea to
check http://www.opengroup.org/onlinepubs/000095399/functions/exec.html

\end{offtopic}

Martien
--
|
Martien Verbruggen | It's not what we don't know that hurts us,
| it's what we know for certain that just ain't
| so. -- Mark Twain

Martien Verbruggen, Jun 27, 2008
12. Antoninus TwinkGuest

On 27 Jun 2008 at 0:48, Martien Verbruggen wrote:
>> On Jun 26, 12:28Â pm, Antoninus Twink <> wrote:
>>> Â  Â  Â  execlp("kedit", "kedit", (char *) NULL);

>>
>> Maybe I misunderstood section 8.10 in the book "Advance Programming in
>> the Unix Environment" by Stevens and Rago, but I believe
>> (char *)NULL actually creates undefined behavior when it's passed to
>> the exec() family.

>
> \begin{offtopic}

?? Are you serious? Surely even the most anal topicality zealot would
accept that how to pass arguments to a variadic function is topical in
clc? I guess not.

> I think you've got it the wrong way around. The last argument in the
> variable list _has_ to be cast to char *.

To elaborate on this for the benefit of "Chad", who prefers to insult
other people rather than learn from them: this isn't just language
fussiness, but is necessary in the real world too.

NULL is a macro, commonly defined either as (void *)0, or just as plain
0. In the first case, there's no problem. In the second case, because by
definition there's no prototype for the arguments in the variable part
of a variadic function, if the preprocessor expands NULL to plain 0,
then the compiler will apply the default promotions and pass 0 as an
int. Things will go wrong in the common situation where int is 32 bits,
but pointers are 64 bits.

Antoninus Twink, Jun 27, 2008
13. Stephen SprunkGuest

Antoninus Twink wrote:
> On 27 Jun 2008 at 0:48, Martien Verbruggen wrote:
>>> On Jun 26, 12:28 pm, Antoninus Twink <> wrote:
>>>> execlp("kedit", "kedit", (char *) NULL);
>>> Maybe I misunderstood section 8.10 in the book "Advance Programming in
>>> the Unix Environment" by Stevens and Rago, but I believe
>>> (char *)NULL actually creates undefined behavior when it's passed to
>>> the exec() family.

>> \begin{offtopic}

>
> ?? Are you serious? Surely even the most anal topicality zealot would
> accept that how to pass arguments to a variadic function is topical in
> clc? I guess not.

No prototype (and other documentation) for execlp() was listed, so in
theory it's off-topic here. If you had said "my system has a variadic
function named execlp(), and the last parameter must be of type (char *)
[or (void *)]", folks not familiar with POSIX would be able to comment.

>> I think you've got it the wrong way around. The last argument in the
>> variable list _has_ to be cast to char *.

>
> To elaborate on this for the benefit of "Chad", who prefers to insult
> other people rather than learn from them: this isn't just language
> fussiness, but is necessary in the real world too.
>
> NULL is a macro, commonly defined either as (void *)0, or just as plain
> 0. In the first case, there's no problem. In the second case, because by
> definition there's no prototype for the arguments in the variable part
> of a variadic function, if the preprocessor expands NULL to plain 0,
> then the compiler will apply the default promotions and pass 0 as an
> int. Things will go wrong in the common situation where int is 32 bits,
> but pointers are 64 bits.

Or in the case of processors that pass pointers and integers in
different registers...

However, I would expect that in either of those cases, the
implementation's definition of NULL would be of the "(void *)0" form,
not the "0" form, specifically to eliminate problems like this. Since
the representations of (void *) and (char *) must be the same, it would
take a rather perverse implementation (e.g. the DS9k) to pass them

S

Stephen Sprunk, Jun 29, 2008
14. Stephen SprunkGuest

Antoninus Twink wrote:
> On 26 Jun 2008 at 19:13, wrote:
>>>> wrote:
>>>>> what i would like to figure out is to make all 5 kedits open at once,

>> Thanks for all the responses sorry for posting here, but the
>> system("kedit &"); did work read the man on system() little bit better
>> understanding of whats going on. i looked up 10 different
>> documentations on fork/exec it seems a little complicated just yet

>
> It's really not all that complicated. Here's a simple example that
>
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <unistd.h>
> #include <sys/types.h>
>
> int main(void)
> {
> pid_t pid;
> int count;
> for(count=0; count<5; count++) {
> pid=fork();
> if (pid==0) {
> /* child process */
> execlp("kedit", "kedit", (char *) NULL);
> perror("exec failure");
> exit(1);
> }
> }
> return 0;
> }
>

<OT>
If you're having trouble figuring out what's going on above, consider
this primitive (and buggy) implementation of system():

int system(const char *command) {
pid_t pid = fork();
if (!pid) {
execlp(command, (char *)NULL);
return -1;
} else {
return waitpid(pid, NULL, 0);
}
}

Since the OP was looking to remove the waitpid() part of system(), the
most logical answer is to use fork()/exec() directly.

Once you wrap your head around the concepts that fork() returns twice
and exec() never returns (assuming everything works properly), the whole
process is rather obvious.
</OT>

S

Stephen Sprunk, Jun 29, 2008
15. CBFalconerGuest

> Antoninus Twink <> wrote:
>

.... snip ...
>
>> It's really not all that complicated. Here's a simple example

>

.... snip twink gubris ...
>
> And off course this is again off topic. So not only does this
> troll fail to understand that this is a C forum, but this person
> writes questionable simple *nix code.

That troll understands very well. Its only objective is to disrupt
the newsgroup. Just PLONK it. To do so, get a real newsreader,
such as Thunderbird, and abandon google.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>

CBFalconer, Jun 29, 2008
16. santoshGuest

CBFalconer wrote:

>> Antoninus Twink <> wrote:
>>

> ... snip ...
>>
>>> It's really not all that complicated. Here's a simple example

>>

> ... snip twink gubris ...

I suppose you meant hubris here?

<snip>

santosh, Jun 29, 2008
17. Walter RobersonGuest

In article <>,
Tony Mc <> wrote:
>On Sun, 29 Jun 2008 08:56:57 +0530, santosh <>
>wrote:

>> I suppose you meant hubris here?

>gubris is the GNU implementation of ISO standard hubris.

Not quite: it is the GNU implementation of something -like-
ISO standard hubris, but with extensions that
violate constraints in the ISO standard hubris, with the
extensions turned on by default (under the excuse that
since gubris does not claim to be ISO standard hubris, the
ISO standard's constraints have no authority over gubris.)
--
What will be your last contribution?
-- Supertramp (Fool's Overture)

Walter Roberson, Jun 30, 2008
18. santoshGuest

Walter Roberson wrote:

> In article <>,
> Tony Mc <> wrote:
>>On Sun, 29 Jun 2008 08:56:57 +0530, santosh <>
>>wrote:

>
>>> I suppose you meant hubris here?

>
>>gubris is the GNU implementation of ISO standard hubris.

>
> Not quite: it is the GNU implementation of something -like-
> ISO standard hubris, but with extensions that
> violate constraints in the ISO standard hubris, with the
> extensions turned on by default (under the excuse that
> since gubris does not claim to be ISO standard hubris, the
> ISO standard's constraints have no authority over gubris.)

Given all this, I'm surprised that CBFalconer endorses implementation
specific extensions in this group, which, as he himself so often
observes, is for ISO standard hubris.

santosh, Jun 30, 2008
19. santoshGuest

Walter Roberson wrote:

> In article <>,
> Tony Mc <> wrote:
>>On Sun, 29 Jun 2008 08:56:57 +0530, santosh <>
>>wrote:

>
>>> I suppose you meant hubris here?

>
>>gubris is the GNU implementation of ISO standard hubris.

>
> Not quite: it is the GNU implementation of something -like-
> ISO standard hubris, but with extensions that
> violate constraints in the ISO standard hubris, with the
> extensions turned on by default (under the excuse that
> since gubris does not claim to be ISO standard hubris, the
> ISO standard's constraints have no authority over gubris.)

Given all this, I'm surprised that CBFalconer endorses implementation
specific extensions in this group, which, as he himself so often
observes, is for ISO standard hubris.

santosh, Jun 30, 2008
20. CBFalconerGuest

santosh wrote:
>

.... snip ...
>
> Given all this, I'm surprised that CBFalconer endorses
> implementation specific extensions in this group, which, as he
> himself so often observes, is for ISO standard hubris.

I suspect you have seriously misinterpreted something. Try making
specific statements.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>