Re: killing a forked process (external programs always detach from child process)
Hi,
I'm trying to launch and external program using fork.
my $pid = fork()
if ($pid == 0){
system (top);
exit (0);
}else{
waitpid ($pid, 0)
}
print External Program died!;
The problem is top, or the program i'm actually going to run always detaches
to the child perl process I have forked. If I open another terminal while my
script is running and kill the child's PID, (usually the one with higher number
because there are two perls running), the external program still keeps running..
How do i tell it to stick to the child process so that when i kill the child,
the external program will die. Is there a way to capture the PID of the
external program? I tried using open to launch it but open quickly exits the
program after assigning it to a variable, e.g my $extern_pid = open TOP, top
| or die $!;
--- On Mon, 4/13/09, Michael Alipio daem0n...@yahoo.com wrote:
From: Michael Alipio daem0n...@yahoo.com
Subject: killing a forked process
To: begginers perl.org beginners@perl.org
Date: Monday, April 13, 2009, 4:49 PM
Hi,
My program looks like this:
my $pid = fork ();
if ($pid == 0){
exec (top);
}else{
my $time = 0
while (1){
sleep 1;
$time++;
if ($time == 10){
kill 9, $pid;
}
}
}
That is, after running the forked top process
for 10 seconds, the main program will end it.
Any idea why it's not working for me?
Thanks
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Re: killing a forked process
On Mon, Apr 13, 2009 at 04:49, Michael Alipio daem0n...@yahoo.com wrote:
snip
my $pid = fork ();
You must check to see if $pid is defined. If it isn't
then fork failed. The parentheses are oddly placed and
unnecessary.
if ($pid == 0){
exec (top);
Again with the odd parentheses placement. Function
calls should look like this
func($blah, $blah);
Control structures should look like this
while (1) {
if ($it_is_true) {
That space is one of the ways humans use to tell
function calls and control structures apart.
}else{
There is no need to make this an else statement, exec
replaces the current process, so the child can never
reach the rest of the code
my $time = 0
You are missing a semi-colon at the end of this statement.
while (1){
You are trying to read from a non-existent filehandle
named 1, I assume you meant to say
while (1) {
to get an infinite loop. This is also why your script
will never end: this while loop has no exit condition
and no calls to last in its body.
sleep 1;
$time++;
if ($time == 10){
Why all of this mess instead of a simple sleep 10?
kill 9, $pid;
SIGKILL is the kill of last resort, try SIGTERM instead.
It is also a good idea to reap the child with waitpid at
this point. Right after this statement is where a call
to last should go.
}
}
}
It might be a good idea to run stty sane to fix the
terminal (top leaves it in a bad state if killed).
#!/usr/bin/perl
use strict;
use warnings;
die could not fork unless defined(my $pid = fork);
if ($pid == 0) {
exec top;
die could not exec top;
}
sleep 3;
kill 15, $pid;
waitpid $pid, 0;
#fix the terminal after killing top
system stty, sane;
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
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