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A script that spawns multiple instance of itself.

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# 1  
Old 01-30-2011
A script that spawns multiple instance of itself.
Hi all,I have some questions of forking new process,the code is below.Any help will be appreciated.
Code:
  1 #! /bin/bash
  2 
  3 PIDS=$(pidof sh $0)
  4 P_array=( $PIDS )
  5 echo $PIDS
  6 let "instances = ${#P_array[*]}-1"
  7 
  8 echo "$instances instance(s)" of this script running."
  9 echo "[Hit Ctl-C to exit.]";echo
 10 
 11 sleep 1
 12 sh $0

On line 3,in my opinion,it should recursive forever because sh execute the script again,the following will never be run,however,the code did exit if i omit the code below line 3.Second question,why subtract 1 on line 6?Thx.
# 2  
Old 01-31-2011
What is this script supposed to do?
Code:
PIDS=$(pidof sh $0)

creates a child process; I do not see any wait command. What is it you want the script to do?

One standard way to create multiple children and wait for them:
Code:
#!/bin/bash
./myscript.sh arg1 &
./myscript.sh arg1 &
./myscript.sh arg1 &
wait

This creates three children, lets them run, and will not exit until all of them have completed.
This User Gave Thanks to jim mcnamara For This Post:
homeboy
# 3  
Old 01-31-2011
Thanks for your reply.My code is a example code from a book.Probably,it wants to show how to use pidof and how "child process" works.What i can't understand is "pidof sh $0",it starts a new process.why didn't the script goes to execute the new process then went into a endless recursive?
# 4  
Old 02-01-2011
pidof is a program that returns pids that are already running. A child process is needed to get pidof running. Then pidof scans the running processes are reports its findings then exits. It does not start anything else. Read the man page for pidof.
This User Gave Thanks to Perderabo For This Post:
homeboy
# 5  
Old 02-01-2011
clear,Thanks all of you!
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