Re: why doesn't this argument list need a comma after the 1st argument?
- From: Ben Morrow <ben@xxxxxxxxxxxx>
- Date: Sun, 11 Mar 2007 03:25:09 +0000
Quoth "Dave Slayton" <evad.notyals@xxxxxxxxx>:
Sorry, another newbie question:
I'm reading this very interesting book on Perl (Effective Perl Programming
by Joseph N. Hall with Randal L. Schwartz), and here on page 110 there's an
example of a
call to a (prototyped) subroutine that requires 3 arguments: a coderef, a
scalar, and an array, and here's the call:
for_n {print "$_[0], $_[1]\n"} 2, @a;
I understand the parentheses around the list of arguments are optional, and
that the anonymous subroutine does not require the "sub" keyword, but what I
don't understand is how the call gets away with not having a comma after the
closing curly brace and before the 2. Can anyone shed some light on this
for me?
It's a special case. Prototypes were introduced to allow you to write
subs that parse like Perl builtins; so, to allow a map-like sub to be
written, a sub with its first argument prototyped '&' will accept a bare
block (without a comma) like this, and treat it as an anon sub.
Actually, this is the only case where the sub keyword is optional: a sub
prototyped as ($&$) would still need to be called like
foo 1, sub {...}, 2;
See perldoc perlsub for all the details.
Ben
--
BEGIN{*(=sub{$,=*)=sub{print@_};local($#,$;,$/)=@_;for(keys%{ #ben@xxxxxxxxxxxx
$#}){/m/&&next;**=${$#}{$_};/(\w):/&&(&(($#.$_,$;.$+,$/),next);$/==\$*&&&)($;.$
_)}};*_=sub{for(@_){$|=(!$|||$_||&)(q) )));&((q:\:\::,q,,,\$_);$_&&&)("\n")}}}_
$J::u::s::t, $a::n::o::t::h::e::r, $P::e::r::l, $h::a::c::k::e::r, $,
.
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