in reply to Re^2: shift vs @_
in thread shift vs @_
Both ways are ok. $_[0] is actually a reference (thanks, tye!) an alias to the first argument passed by the caller. You can evaluate (or operate on) $_[0] directly. Passing arguments as references just adds another level of indirection.
You can also use prototypes for your subs and access your arguments inside the sub as references:
#!/usr/bin/perl sub foo(\@\$) { warn "foo args: (".join(",",map{"'$_'"}@_).")\n"; print "1st argument = $_[0]; content =(" . join(',',map{"'$_'"} @{$_[0]}).")\n"; print "2nd argument = $_[1]; content = '".${$_[1]}."'\n"; my ($array,$scalar) = @_; push @$array, $$scalar; $$scalar = "blurf"; } my @ary = qw(foo bar baz); my $foo = "blah"; foo(@ary,$foo); print "ary: (@ary)\n"; print "foo: $foo\n";
output:
foo args: ('ARRAY(0x8167870)','SCALAR(0x8167978)') 1st argument = ARRAY(0x8167870); content =('foo','bar','baz') 2nd argument = SCALAR(0x8167978); content = 'blah' ary: (foo bar baz blah) foo: blurf
--shmem
_($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
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Re^4: shift vs @_ (alias)
by tye (Sage) on Oct 03, 2006 at 21:26 UTC |
In Section
Seekers of Perl Wisdom