not getting the correct Process ID

culin · 05-02-2007, 02:18 AM

Hi all,
below is the piece of code i am working on...i have some doubts regarding this....what i wanted from this code is, it has to print the process ID of the vsftpd that is running ( which is started by me by doing execvp(); )

Code:

#include<stdio.h>
#include<unistd.h>
#include<stdlib.h>
#include<string.h>
#include<sys/types.h>
int main()
{
        char *arg[]={ "VSFTPD",
                        "restart",
                        NULL  };
        int child_pid,*status;
        child_pid=fork();
        if (child_pid == 0)
        {
                //sleep(5);
                execvp("/etc/init.d/vsftpd", arg);
                fprintf (stderr, "\n an error occurred in execvp\n\n\n");
        }
        else
        {
                sleep(5);
                printf("the child_pid is %d\n",child_pid);
        }
        return 0;
}

but the problem what is have is when i do an execvp() inside the child it is starting 2 processes named vsftpd one is /usr/sbin/vsftpf(the pid of which i am expecting to get printed ) and the other is <defunct vsftpd>... but it is printing the pid of the <defunct vsftpd> which will not be in the process list after 5 seconds... so how do i get the pid of the /usr/sbin/vsfptd by using this piece of code...??? please gimme ur suggestions.. and please tell me why this <defunct vsftpd> is coming.....???

output of my program..

Code:

[root@localhost ~]# ./a.out
Shutting down vsftpd:                                      [  OK  ]
Starting vsftpd for vsftpd:                                [  OK  ]
the child_pid is 5560

the process list

Code:

root      5559  0.0  0.0   1408   268 pts/5    S+   18:09   0:00 ./a.out
root      5560  0.0  0.0      0     0 pts/5    Z+   18:09   0:00 [vsf] <defunct>
root      5570  0.0  0.2   4064  1076 ?        S    18:09   0:00 /usr/sbin/vsftp
root      5573  0.0  0.1   4488   924 pts/4    R+   18:09   0:00 ps -aux

THANKS....

vazrik · 05-02-2007, 02:36 AM

To avoid the defunct (zombie) process, you have to wait for the child to finish. Lookup wait and waitpid man pages.

It seems that you're successfully executing the script to bring down ftpd and bring it up again, which it does.

Hope that helps.

culin · 05-02-2007, 03:30 AM

thanks for reply vazrik,
but as i know, execvp() is a blocking call it wont execute the next statement untill there is a failure in execvp()..

vazrik · 05-03-2007, 04:28 AM

I hope you've solved your problem by now, but just in a remote case ...

Your exec() (and variations of it) creates a child process, and initiates it, then it returns. From that point on, the child process and the parent process run asynchroneusly (sp?) with respect to each other. Therefore, your parent process (just before the return) must wait for the child process to finish in order to not create a zombie. I am a little foggy on these, since I have not done these for some years, but that is basically the jist of it. I am foggy in particular about the initiation part. The best I remember (from Solaris), I think it sets it to runnable and then returns. At which point the parent continues executing to the end of its time-slice or

At any rate, if the parent exits (terminates) before the child, then the child process becomes zombie.

theNbomr · 05-03-2007, 05:07 PM

Actually, exec() family calls do not return. They simply exit. You are correct that you must call wait() on those child processes to prevent that nasty 'zombie buildup'.

--- rod.

vazrik · 05-03-2007, 06:18 PM

Hi Rod,

Thank you for "making" me read and refresh my memory :-)

You are absolutely right about exec() not returning. It overrides/replaces the running process memory image with that of the new code.

However, since you do fork() and then exec() in the child process, the parent process continues to run as well. You execute the exec() in the child process after forking.

I wonder, if you don't fork at all, would you still get the zombies? According to my understanding, you should not.