Re: #!/Perl question
Hi,
>>"Joey" == Joey Hess <joey@kitenet.net> writes:
>> The following works under more shells (and also is a man page ;-)
#!/usr/bin/perl -- # -*- Mode: Perl -*- #
'di';
'ig00';
"true" || eval 'exec perl -S $0 $argv:q';
eval '(exit $?0)' && eval 'exec perl -S $0 ${1+"$@"}'
& eval 'exec perl -S $0 $argv:q'
if $running_under_some_shell;
Joey> Yes, but it doesn't address the orginial problem: It will fail if perl is
Joey> not in /usr/bin.
Well. Just get rid of the #! line, and it works as expected (I
really shouyld have caught that).
See, if not invoked under Perl, the first two lines are shell
no-ops, and the third line execs Perl.
I still use parts of the template.
I remember seeing a program that was legel in 7 languages ...
manoj
--
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Manoj Srivastava <srivasta@debian.org> <http://www.debian.org/%7Esrivasta/>
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