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

If the child exits and the parent is still alive, then the system keeps an association of child with parent, as the exit code needs to be stored in case the parent subsequently calls a wait. The child becomes zombie till the parent does a wait.
however if the parent has exited before the child exiting, then the child is adopted by the init which reaps the exit code and the child does not become a zombie.