Interesting point, especially if operator:+= can be overloaded. @a [+=] @b;
implies iteratively invoking operator:+=, whereas @a [+]= @b; implies assigning the result of iteratively invoking operator:+ It only matters when they're different. :-| And, of course, if they ARE different then the implication is that signature-matching may also be involved, such that the version of operator:+ (or +=) that gets used can change... Hell, we might as well throw in multiple dispatch. Any of you OO guys know of a case where $a = $a + $b; # @A [+]= @B; --> @A = @A [+] @B; and $a += $b; # @A [+=] @B; should be different? (Intuitively, I get the feeling that these cases exist, especially in the weird cases kind of like the >> and << operators that C++ redefined for I/O. But I'm not sure if that's paranoia causing me to accept lousy design, or what...) Or should there be an overloadable operator:[op]= that takes arrays or lists as its normal context? (Kind of like C++ differentiating between new and new[]) =Austin --- Luke Palmer <[EMAIL PROTECTED]> wrote: > > Mailing-List: contact [EMAIL PROTECTED]; run by ezmlm > > Date: Tue, 29 Oct 2002 21:37:32 +0000 > > From: Aaron Crane <[EMAIL PROTECTED]> > > Content-Disposition: inline > > X-SMTPD: qpsmtpd/0.12, http://develooper.com/code/qpsmtpd/ > > > > Damian Conway writes: > > > My personal favorite solution is to use square brackets (for > their dual > > > array and indexing connotations, and because they highlight the > operator > > > so nicely): > > > > > > $count = @a + @b; > > > @sums = @a [+] @b; > > > > Mmm, yummy. I do have a question though (and apologies if I've > merely > > missed the answer). We've got two productive operation-formation > rules: one > > saying "add a final = to operate-and-assign", and the other saying > "wrap in > > [] to vectorise". But no-one's said which order they apply in. > That is, > > which of these should I type: > > > > @x [+]= @y; > > @x [+=] @y; > > > > Of course, the rule ordering didn't matter with the "add a leading > ^ to > > hype" rule. > > Hmm. Well, they're different: > > @x [+]= @y; > @x = @x [+] @y; > > @x [+=] @y; > for @x | @y -> $x is rw | $y { > $x += $y > } > > :) Is there any advantage in differentiating that? Or would the > former *always* optimize to the latter? > > Luke > > > I think I prefer the first one, by the way -- it strikes me as more > > obviously a vector add. > > > > -- > > Aaron Crane * GBdirect Ltd. > > http://training.gbdirect.co.uk/courses/perl/ __________________________________________________ Do you Yahoo!? HotJobs - Search new jobs daily now http://hotjobs.yahoo.com/
