On Sat, Jan 28, 2006 at 06:30:06PM +0100, dakkar wrote:
: Today I was reading S06 from
: http://dev.perl.org/perl6/doc/design/syn/S06.html, and I have some
: perplexities on what's written there.
:
: 1) it's written:
:
: $pair = :when<now>;
: doit $pair,1,2,3; # always a positional arg
: doit *$pair,1,2,3; # always a named arg
:
: but also:
:
: To pass pairs out of hash without their being interpreted as named
: parameters, use
:
: doit %hash<a>:p,1,2,3;
: doit %hash{'b'}:p,1,2,3;
:
: instead.
:
: I interpret the last sentence as meaning that:
:
: $hash<a>=:a<2>; doit %hash<a>;
:
: would be equivalent to:
:
: doit :a<2>
:
: which contradicts what is written earlier... I'd assume I'd have to
: write:
:
: doit *(%hash<a>);
:
: for the pair to be used as a named argument. What am I missing?
You're reading "pass pairs" as meaning to return a particular value
that happens to be a pair. That's not the purpose of :p, since that's
what happens anyway by default. You use :p to return both the key
and the value as a pair object from the hash. If you say:
%hash<b>=:a<2>; doit %hash<b>;
it's equivalent to
doit (:a<2>)
whereas if you say
%hash<b>=:a<2>; doit %hash<b>:p;
it is equivalent to
doit (:b(:a<2>))
In neither case is the argument taken as a named argument.
I have rewritten the paragraph as:
Ordinary hash notation will just pass the value of the hash entry as a
positional argument regardless of whether it is a pair or not.
To pass both key and value out of hash as a positional pair, use C<:p>.
doit %hash<a>:p,1,2,3;
doit %hash{'b'}:p,1,2,3;
instead. (The C<:p> stands for "pairs", not "positional"--the
C<:p> adverb may be placed on any hash reference to make it mean
"pairs" instead of "values".)
to try to make it clearer.
: 2) Consider:
:
: doit %*hash; # 1 arg, the global hash
: doit *%hash; # n args, the key-value pairs in the hash
:
: I think I'll use a lot of C-t while writing Perl6 code...
The * as shorthand for GLOBAL:: is allowed only where it wouldn't be confused
with unary *. That is, only in declarative contexts or as a twigil.
: 3) What does the second line in:
:
: push @array, :a<1>;
: say *pop(@array);
:
: mean? I'd parse it as 'say (*pop)(@array)', that is, call the global
: 'pop' subroutine. But the text around this example seems to imply that
: it is to be parsed as 'say *(pop(@array))', i.e. execute 'pop' and splat
: the result. What gives?
The above rule. If we meant the global pop we'd say:
say &*pop(@array);
or
say GLOBAL::pop(@array);
Gotto go to lunch--more in a bit. (Please use new subjects if you want
to discuss any of these subpoints further...)
Larry
