Oh look, this isn't an argument. update2: video

sorry, couldn't resist ;-)

update: 

foo($bar, $baz); 
# ...
sub foo{        
     print "$_";   
#will first print out the contents of $bar, then $baz 
...

[download]
No it won't.

you mean 

sub foo {
  print for @_;
}
# or
sub foo {
  print @_;
}

[download]
"What should it profit a man, if he should win a flame war, yet lose his cool?"

In reply to Re: Tutorial on arguments for the new by Joost
in thread Tutorial on arguments for the new by Andrew_Levenson

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