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perl opening code line - need explanation:

Hello,

i've encountered a script which has the following code line on top:

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  1. eval '(exit $?0)' && eval 'exec perl -w -S $0 ${1+"$@"}' & eval 'exec perl -w -S $0 $argv:q'
  2.  

can anybody explain me what does it says. the only initials i've seen so far are !#/usr/bin/perl, which tells the server where to look for the perl interpreter .

thanks,
Jun 8 '08 #1
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numberwhun
Expert Mod 2GB
Not knowing whether this is homework you are supposed to work out or not, I will provide you with a couple of places you can start looking.

First, the command line options to perl can be found on the perlrun page. That site, perldoc.perl.org, is also an excellent reference for the language.

Also, in the beginning, there is a "$?0". To me, this makes no sense because the $? is the 16 bit status word returned by the last wait or waitpid syscall. I don't think the zero should be there.

My suggestion, since you are learning, is to figure this out. You should be able to figure out what it is doing using the perldoc page, among others.

Regards,

Jeff
Jun 9 '08 #2

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