This is probably blindingly simple but I'm not understanding why @ARGV is not reduced to () (no args)? In this example,I am running the command as below which ends up as 6 arguments but why $#ARGV !=6 is true? 

I am running the command using the following

perl perl.pl -b opt_b -p opt_p -c opt_c

if ($#ARGV == -1 || $#ARGV !=6) {
print "ARGV is now reduced to <".@ARGV."> members:\n";#prints ARGV as 
+6
}

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In reply to ARGV behaviour in getopts std by perl_mystery

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