lo_tech has asked for the wisdom of the Perl Monks concerning the following question:
Could anyone explain this to me?
Here's the code:
#!/usr/bin/perl -Tw use strict; print $#ARGV; die "where are those args again?\n" unless ($#ARGV > 0); exit;
Here's the output without an @ARGV:
where are those args again?<br> -1
Here's the question:
Why is the print after the die?
Server Error (Error ID 921132c2133309)
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Replies are listed 'Best First'. | |
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Re: Does die() change the order of execution?
by runrig (Abbot) on Nov 01, 2001 at 21:50 UTC | |
Re: Does die() change the order of execution?
by jwest (Friar) on Nov 01, 2001 at 21:53 UTC | |
Re: Does die() change the order of execution?
by mkmcconn (Chaplain) on Nov 01, 2001 at 22:00 UTC | |
Re: Does die() change the order of execution?
by Asim (Hermit) on Nov 01, 2001 at 21:58 UTC |
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