On Thu, Mar 11, 2004 at 02:05:55PM -0600, Jonathan Scott Duff wrote:
: On Thu, Mar 11, 2004 at 11:11:54AM -0800, Larry Wall wrote:
: > On the final hand, if people fall in love with both self:sort and =sort, we
: > could have =sort be a shorthand for self:sort where it's unambiguous.
:
: Wouldn't =sort potentially muck with POD?
Could. Historically pod only pays attention to = on the left margin though.
So you generally wouldn't have any problem unless you were in the habit
of declaring your methods in the C-ish idiom of:
int method
=rotate (int $a is rw) {...}
On the other hand, I suspect most people will end up declaring it
int method
self:rotate (int $a is rw) {...}
in any event, and reserve the =rotate for .=rotate, which can never put
the = on the left margin, even if we let ourselves have whitespace
before POD directives. So maybe we just require self: for the declaration,
and forget about = there. It interacts badly with global names anyway.
Is it "*=sort" or "=*sort"? With "*self:sort" it's more obvious.
Another interesting question, if the "postfix:.=foo" mess is defined
with as self:foo, should infix:+= be defined as self:+ instead?
In other words, should the <op>= syntax really be a metasyntax like
hyperoperators, where you never actually have to define a C<»+«>
operator, but the hyperoperator is always autogenerated from ordinary
C<+>? So basically any infix:<op>= gets remapped to self:<op>.
In that case, C<»+=«> is a double-meta operator that ends up generating
a hyper self:+.
I kinda like this approach because it means you can always get all of
$a !! $b
$a !!= $b
@a »!!« @b
@a »!!=« @b
merely by defining infix:!!. On the other hand, it also means that
someone can say silly things like:
$a cmp= $b
$a ~~= $b
I suppose we could simply disallow meta-= on non-associating operators.
Can anyone come up with a non-associating binary operator that *should*
have an assignment operator? The basic definition of non-associating
seems to be that the type of the arguments is incompatible with the
type produced by the operator. Which is precisely the problem with
something like
$a cmp= $b
insofar as $a is being treated as a string at one moment and as a boolean
at the next.
Larry
