This is probably blindingly simple but I'm not understanding why @ARGV
is not reduced to () (no args)? in this example
I'm trying to understand what happens with ARGV in s getops loop with
arguments after the getopts Options, or any time there is ARGV, I
guess.
Why doesn't the `for loop' reduce ARGV to `()'?
The script included at the end, when given these arguments:
`./test.pl -a A -b B one two three'
Generates this output:
ARGV starts with <7> members
$opt_a is <A>
$opt_b is <B>
ARGV is now reduced to <3> members:
one
two
three
Now lets ditch the rest in a for loop
Ditching <one>
Ditching <two>
ARGV after shifting away args in `for' loop:
ARGV=<three>
OOOPS... ARGV still has <1> member
------- --------- ---=--- --------- --------
#!/usr/local/bin/perl
use strict;
use warnings;
print "ARGV starts with <"....@argv."> members\n";
## ==== BEGIN Getopts section ====
## Declare vars inside our()
our ($opt_a,$opt_b);
use Getopt::Std;
my $optstr ="a:b:";
getopts($optstr);
if ($opt_a){
print "\$opt_a is <$opt_a>\n";
}
if ($opt_b){
print "\$opt_b is <$opt_b>\n";
}
print "ARGV is now reduced to <"....@argv."> members:\n";
for (@ARGV){
print "$_\n";
}
print "Now lets ditch the rest in a for loop\n";
for (@ARGV){
my $ditch = shift;
print "Ditching <$ditch>\n";
$ditch = '';
}
print "ARGV after shifting away args in \`for' loop:
ARGV=<@ARGV>
OOOPS... ARGV still has <". @ARGV ."> member\n";
@ARGV = ();
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