UNIX for Advanced & Expert Users

Hi all,

Please look into the following code :

int main()
{

char command[ 1024 ];
int pid, ppid;

ppid = getpid(); /* Get the parent pid */

pid = fork(); /* Fork */

if ( pid ==0 )
{

sprintf( command, " gdb a.out %d ", ppid );
printf( "Command line is %s\n", command );

system( command );

}

else if ( pid > 0 )
{

int x=1;
printf(" In parent \n" );

while( x );
printf ( "Exiting parent \n" );

}

return 0;

}

As you note, I'm trying to debug the parent from the child process.
I compiled the above program and tried running it on a linux machine. I got the gdb prompt and issued the command set x=0 from gdb prompt, so that the parent comes out of the infinite loop, and then I pressed continue in gdb. The parent exited normally in gdb. The strange fact is that the child also has exited normally without being a zombie. How could this happen ?

I tried this for many times, still I find no zombies on my machine . Can anyone please explain this behaviour

Bye

i guess the value you are setting to 'x' is having different scope
from the 'x' of while loop of parent process.

If a parent dies, the child process is adopted by init. When the child dies, init will reap it. To create a zombie, the child must die and the living parent must not issue a wait.

This message has been deleted