HaloO,
Jon Lang wrote:
Actually, note that both infix:<,> and circumfix:<[ ]> can be used to
build lists; so [1] and [] can be used to construct single-element and
empty lists, respectively.
I doubt that. Actually, circumfix:<[ ]> builds arrays. And note that
there's no infix operator that concatenates arrays. '[1,2],[3,4]' is
a two element list. The closest you can get is '[1,2].push: [3,4]'
which preserves the identity of the first array whereas the lists
in infix<,> behave like values. I.e. the concatenated list has got
a new identity, not that of the left list.
Personally, I'd like to see '()' capture the concept of "nothing" in
the same way that '*' captures the concept of "whatever". There _may_
even be justification for differentiating between this and "something
that is undefined" (which 'undef' covers). Or not; I'm not sure of
the intricacies of this. One possibility might be that '1, 2, undef'
results in a three-item list '[1, 2, undef]', whereas '1, 2, ()'
results in a two-item list '[1, 2]' - but that may be a can of worms
that we don't want to open.
I see no can of worms. A *defined* () as "nothing" is just as well
defined as 0 is for addition of numbers or '' for string concatenation.
The thing that Larry's line of thought leads to is that
my $a = () // (1,2);
means that $x receives (1,2) not the "nothing" object. The question
that arises when comparing () to * is how polymorph () is:
my $b = 3 + (); # $b == 3?
my $c = 'ab' ~ (); # $c eq 'ab'?
my $d = 1,2 , (); # $d === (1,2)
my $e = 3 * (); # $e == 0?
my $f = 3 ** (); # $f == 1?
my $g = () ?? 1 !! 2; # $g == 2?
my $h; # $h === ()? scalar defaults to nothing
Do the Nothing and Whatever have unique IDs?
my %h;
%h{()} = "Nothing";
%h{*} = "Whatever";
%h{*+3} = "Whatever plus three";
I guess for hashes there is no slicing implied by the latter two cases,
so these would use the .WHICH to build the index.
Regards, TSa.
--
"The unavoidable price of reliability is simplicity" -- C.A.R. Hoare
"Simplicity does not precede complexity, but follows it." -- A.J. Perlis
1 + 2 + 3 + 4 + ... = -1/12 -- Srinivasa Ramanujan
